Moving Charges and Magnetism: NEET and JEE questions with solutions

Answer: 20.9

The magnetic force provides the centripetal force: $qvB = \dfrac{mv^2}{r}$.
$$r = \frac{mv}{qB} = \frac{(1.67\times10^{-27})(2\times10^{6})}{(1.6\times10^{-19})(0.1)} \approx 0.209\ \text{m} = 20.9\ \text{cm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{\mu_0 I_0}{12r}$

The current enters at A and leaves at C, which are diametrically opposite. The straight lead wires lie along the line through O, so they give no field at O.

The two semicircles ABC and ADC are in parallel. Since $R_{ABC} = \tfrac{1}{2}R_{ADC}$, the current divides as
$$I_{ABC} = \frac{2I_0}{3}, \qquad I_{ADC} = \frac{I_0}{3}$$

A semicircle carrying current $I$ gives $B = \dfrac{\mu_0 I}{4r}$ at the centre. The two halves carry current in opposite senses around O, so their fields oppose:
$$B = \frac{\mu_0}{4r}\left(\frac{2I_0}{3} - \frac{I_0}{3}\right) = \frac{\mu_0 I_0}{12r}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{\mu_0 I^2}{\pi\lambda g}$

Currents in the same direction attract, so wire A is pulled upward towards B. Force per unit length on A:
$$\frac{F}{l} = \frac{\mu_0 (I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$$

A does not rise as long as this does not exceed its weight per unit length $\lambda g$:
$$\frac{\mu_0 I^2}{\pi h} \le \lambda g \;\Rightarrow\; h \ge \frac{\mu_0 I^2}{\pi\lambda g}$$

Minimum $h = \dfrac{\mu_0 I^2}{\pi\lambda g}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^2\ \text{V m}^{-1}$

For no deflection, the electric force must cancel the magnetic force: $q\vec{E} = -q\vec{v} \times \vec{B}$. The magnetic force is perpendicular to $\vec{B}$, so $\vec{E}$ must be perpendicular to $\vec{B}$ (and to $\vec{v}$).

Magnitude: $eE = evB$, so
$$E = vB = \frac{3 \times 10^8}{100} \times 9 \times 10^{-4} = 3 \times 10^6 \times 9 \times 10^{-4} = 27 \times 10^2\ \text{V m}^{-1}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $1:4$

Magnetic moment of a coil: $m = NIA = NI\pi r^2$. With the same current (and the same number of turns),
$$\frac{m_1}{m_2} = \frac{r_1^2}{r_2^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{he}{2\pi m}$

Lowest state, $n = 1$: $B\pi r^2 = \dfrac{h}{e}$, so $r^2 = \dfrac{h}{\pi eB}$.

In the magnetic field, $r = \dfrac{mv}{eB}$, so $v = \dfrac{eBr}{m}$.

Magnetic moment of a circulating charge:
$$\mu = IA = \frac{ev}{2\pi r}\cdot\pi r^2 = \frac{evr}{2} = \frac{e^2Br^2}{2m}$$

Substituting $r^2$:
$$\mu = \frac{e^2B}{2m}\cdot\frac{h}{\pi eB} = \frac{he}{2\pi m}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4.4$ mT

$$B = \frac{\mu_0 NI}{2R} = \frac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 0.1} = 4.4 \times 10^{-3}\ \text{T} = 4.4\ \text{mT}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $5\,IL$

$$\vec{F} = I\vec{L} \times \vec{B} = IL\,\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k}) = IL(3\hat{k} + 4\hat{j})$$

(using $\hat{i}\times\hat{i} = 0$, $\hat{i}\times\hat{j} = \hat{k}$, $\hat{i}\times\hat{k} = -\hat{j}$).

$$|\vec{F}| = IL\sqrt{3^2 + 4^2} = 5\,IL$$

(The component of $\vec{B}$ along the wire, $2\hat{i}$, exerts no force.)

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{\mu_0 i}{4R}\left[1 - \dfrac{2}{\pi}\right]$ pointed away from page

P is the centre of the semicircle AB of radius $R$. The wire has three parts.

Semicircle: the current goes from A (top) round the left side to B (bottom), which is anticlockwise as seen, so its field at P points out of the page:
$$B_1 = \frac{\mu_0 i}{4R}$$

Upper straight wire: semi-infinite, ending at A, directly above P at distance $R$. The current flows to the right and P is below the wire, so the field is into the page:
$$B_2 = \frac{\mu_0 i}{4\pi R}$$

Lower straight wire: semi-infinite, ending at B. The current flows to the left and P is above the wire, so the field is also into the page:
$$B_3 = \frac{\mu_0 i}{4\pi R}$$

Net field (out of the page positive):
$$B = \frac{\mu_0 i}{4R} - \frac{2\mu_0 i}{4\pi R} = \frac{\mu_0 i}{4R}\left[1 - \frac{2}{\pi}\right]$$
pointing away from (out of) the page.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) see figure

Ampère's law with a circle of radius $r$:

Inside ($r < R$), the enclosed current is $I\dfrac{r^2}{R^2}$:
$$B \cdot 2\pi r = \mu_0 I\frac{r^2}{R^2} \;\Rightarrow\; B = \frac{\mu_0 I r}{2\pi R^2} \propto r$$

Outside ($r > R$):
$$B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$$

$B$ rises linearly from zero to a maximum at $r = R$, then falls as $1/r$. This is option (4).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $8 \times 10^{-20}$ N

Field of the wire at $20$ cm:
$$B = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 5}{0.2} = 5 \times 10^{-6}\ \text{T}$$

The electron moves parallel to the wire, perpendicular to $B$:
$$F = evB = 1.6 \times 10^{-19} \times 10^5 \times 5 \times 10^{-6} = 8 \times 10^{-20}\ \text{N}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $-6\hat{i} - 6\hat{j} - 8\hat{k}$

$$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 6 \\ B & B & B_0 \end{vmatrix} = (4B_0 - 6B)\hat{i} + (6B - 2B_0)\hat{j} + (2B - 4B)\hat{k}$$

Compare with $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$:

$\hat{k}$: $-2B = 12$, so $B = -6$.

$\hat{i}$: $4B_0 + 36 = 4$, so $B_0 = -8$.

Check $\hat{j}$: $6(-6) - 2(-8) = -36 + 16 = -20$. ✓

$$\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\sqrt{3}Ia^2$ and $3Ia^2$

Magnetic moment: $m = NIA$.

(i) Triangle: perimeter $3a$, so $N = \dfrac{12a}{3a} = 4$ turns, each of area $\dfrac{\sqrt{3}}{4}a^2$:
$$m_1 = 4 \times I \times \frac{\sqrt{3}}{4}a^2 = \sqrt{3}Ia^2$$

(ii) Square: perimeter $4a$, so $N = 3$ turns, each of area $a^2$:
$$m_2 = 3Ia^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $6.28 \times 10^{-4}$ T

Turns per unit length: $n = \dfrac{100}{0.5} = 200\ \text{m}^{-1}$.

$$B = \mu_0 nI = 4\pi \times 10^{-7} \times 200 \times 2.5 = 2\pi \times 10^{-4} \approx 6.28 \times 10^{-4}\ \text{T}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4.9$ cm

Only the component along $B$ carries the proton forward: $v_\parallel = v\cos 60^\circ = 1.5 \times 10^5$ m/s.

Time for one turn: $T = \dfrac{2\pi m}{eB}$.

Pitch $= v_\parallel T = \dfrac{2\pi m v_\parallel}{eB} = \dfrac{2\pi \times 1.67 \times 10^{-27} \times 1.5 \times 10^5}{1.6 \times 10^{-19} \times 0.2} \approx 0.049$ m $= 4.9$ cm.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1 : 1$

$r = \dfrac{mv}{qB} = \dfrac{\sqrt{2mK}}{qB}$.

Proton: $\dfrac{\sqrt{2mK}}{eB}$. Alpha (mass $4m$, charge $2e$): $\dfrac{\sqrt{8mK}}{2eB} = \dfrac{\sqrt{2mK}}{eB}$.

The radii are equal, so the ratio is $1 : 1$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1 : \sqrt{2} : \sqrt{2}$

After acceleration $K = qV$, so $r = \dfrac{\sqrt{2mK}}{qB} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}} \propto \sqrt{\dfrac{m}{q}}$.

Proton $\sqrt{m/e}$; deuteron $\sqrt{2m/e}$; alpha $\sqrt{4m/2e} = \sqrt{2m/e}$.

Ratio $1 : \sqrt{2} : \sqrt{2}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 12

The protons leave when their orbit radius reaches $R$: $v = \dfrac{eBR}{m}$.

$$K = \frac{e^2B^2R^2}{2m} = \frac{(1.6 \times 10^{-19})^2 \times 1 \times 0.25}{2 \times 1.67 \times 10^{-27}} \approx 1.92 \times 10^{-12}\ \text{J}$$

In MeV: $\dfrac{1.92 \times 10^{-12}}{1.6 \times 10^{-13}} \approx 12$ MeV.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1.5 \times 10^6$ m/s

Undeflected means the electric and magnetic forces cancel: $qE = qvB$.

$v = \dfrac{E}{B} = \dfrac{3 \times 10^4}{0.02} = 1.5 \times 10^6$ m/s. The result does not depend on the charge or mass.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $30^\circ$

Inside the field the particle moves on a circular arc of radius $r$. Crossing a strip of width $d$, the arc turns through angle $\theta$ with $\sin\theta = \dfrac{d}{r} = \dfrac{d}{2d} = \dfrac{1}{2}$.

So $\theta = 30^\circ$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) remains the same

The time for a semicircle is half the period: $t = \dfrac{\pi m}{qB}$.

This does not depend on speed. A faster particle moves on a larger semicircle, and the longer path exactly balances its higher speed.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $8 \times 10^{-4}$ N, attractive

$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} = \dfrac{2 \times 10^{-7} \times 10 \times 20}{0.05} = 8 \times 10^{-4}$ N/m.

Currents in the same direction attract.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{2\sqrt{2}\,\mu_0 I}{\pi a}$

Each side is a finite wire at perpendicular distance $\dfrac{a}{2}$ from the centre, and its ends subtend $45^\circ$ on either side:
$$B_1 = \frac{\mu_0 I}{4\pi (a/2)}(\sin 45^\circ + \sin 45^\circ) = \frac{\sqrt{2}\,\mu_0 I}{2\pi a}$$

All four sides give fields in the same direction:
$$B = 4B_1 = \frac{2\sqrt{2}\,\mu_0 I}{\pi a}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{\mu_0 I}{8R}$

The two straight parts point along radii, so their lines pass through the centre and give no field there.

An arc subtending angle $\theta$ gives $B = \dfrac{\mu_0 I}{4\pi R}\theta$. For a quarter circle, $\theta = \dfrac{\pi}{2}$:
$$B = \frac{\mu_0 I}{8R}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{\mu_0 I}{4R}$

Inner loop: $\dfrac{\mu_0 I}{2R}$. Outer loop: $\dfrac{\mu_0 I}{2(2R)} = \dfrac{\mu_0 I}{4R}$, in the opposite direction.

Net: $\dfrac{\mu_0 I}{2R} - \dfrac{\mu_0 I}{4R} = \dfrac{\mu_0 I}{4R}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{1}{8}$

$$\frac{B_{\text{axis}}}{B_{\text{centre}}} = \frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{R^3}{(4R^2)^{3/2}} = \frac{R^3}{8R^3} = \frac{1}{8}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $2 \times 10^{-3}$ T

By Ampere's law around the circular core: $B \cdot 2\pi r = \mu_0 N I$.
$$B = \frac{\mu_0 N I}{2\pi r} = \frac{2 \times 10^{-7} \times 500 \times 2}{0.1} = 2 \times 10^{-3}\ \text{T}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $0.17$ N m

The torque depends on the angle between the field and the coil's normal (its magnetic moment). If the plane is at $30^\circ$ to $B$, the normal is at $60^\circ$.

$\tau = NIAB\sin 60^\circ = 50 \times 2 \times 4 \times 10^{-3} \times 0.5 \times 0.866 \approx 0.17$ N m.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{q\omega R^2}{2}$

The charge passes any point $\dfrac{\omega}{2\pi}$ times per second, so the ring is a current $I = \dfrac{q\omega}{2\pi}$.

$M = I \times \pi R^2 = \dfrac{q\omega}{2\pi} \times \pi R^2 = \dfrac{q\omega R^2}{2}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 1.2

Potential energy $U = -MB\cos\theta$. Stable: $\theta = 0$, $U = -MB$. Unstable: $\theta = 180^\circ$, $U = +MB$.

$W = 2MB = 2 \times 2 \times 0.3 = 1.2$ J.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.05\ \Omega$ in parallel

The shunt must carry the rest of the current, $5 - 0.005 = 4.995$ A, at the same voltage as the galvanometer:
$$S = \frac{I_g G}{I - I_g} = \frac{0.005 \times 50}{4.995} \approx 0.05\ \Omega$$

A shunt is connected in parallel.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $9900\ \Omega$ must be connected in series

At full scale the total resistance must be $\dfrac{10}{0.001} = 10\,000\ \Omega$.

The galvanometer supplies $100\ \Omega$, so add $9900\ \Omega$ in series.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) current sensitivity doubles; voltage sensitivity is unchanged

Current sensitivity $\dfrac{\phi}{I} = \dfrac{NAB}{k}$, so it doubles with $N$.

Voltage sensitivity $\dfrac{\phi}{V} = \dfrac{NAB}{kR}$. Both $N$ and $R$ double, so it is unchanged.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2BIR$

In a uniform field, the force on a curved wire equals the force on the straight wire joining its ends. Here the ends are a diameter, $2R$, apart and the field is perpendicular to it.

$F = BI(2R) = 2BIR$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $5024 \times 10^{-7}\hat{j}$

Concept. $\vec\tau=\vec m\times\vec B$ with $\vec m=IA\hat n$.

Step 1. Moment. $\vec m=100\sqrt2\times\pi\times4\times10^{-4}\times\dfrac{\hat i+\hat k}{\sqrt2}=0.1256(\hat i+\hat k)$.

Step 2. Torque. $0.1256\times4\times10^{-3}\,[(\hat i+\hat k)\times(3\hat i+2\hat k)]=5.024\times10^{-4}\,(3-2)\hat j$.

Answer. $5024\times10^{-7}\hat j$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. Helix along $\vec B=B\hat k$: the $\hat k$ component of velocity is unchanged, and each turn takes $T=\dfrac{2\pi m}{qB}$.

Step 1. Period. $T=\dfrac{2\pi\times5\times10^{-6}}{5\pi\times10^{-6}\times0.1}=20$ s.

Step 2. Distance. $5T=100$ s; $2\times10^{-2}\times100=2$ m.

Answer. $\alpha=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $300$

Concept. Straight wire: $B=\dfrac{\mu_0I}{2\pi r}$. Opposite currents give fields in the same direction between the wires, so they add.

Step 1. Each. $\dfrac{2\times10^{-7}\times30}{0.04}=1.5\times10^{-4}$ T.

Step 2. Total. $3\times10^{-4}$ T $=300\ \mu$T.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 50

Concept. Two conditions on the same galvanometer: as an ammeter $I_gG=(I-I_g)S$; as a voltmeter $I_g(G+R)=V$.

Step 1. Ammeter. $I_gG=(0.5-I_g)\times2\Rightarrow I_g=\dfrac{1}{G+2}$.

Step 2. Voltmeter. $\dfrac{G+470}{G+2}=10\Rightarrow9G=450$.

Answer. $G=50\ \Omega$.

Solution by Sreeraj P, M.Sc Physics

Answer: 12

Concept. The magnetic force (and hence the acceleration) is always perpendicular to $\vec B$: $\vec a\cdot\vec B=0$.

Step 1. Dot product. $4\times3-\dfrac x2\times2=0$.

Answer. $x=12$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{qE_0}{\lambda}\left[1 - e^{-\lambda L}\right]$

Concept. The magnetic force does no work, so the change in kinetic energy is the work done by $\vec E$ alone. $\vec E$ is along $x$, so only the change in $x$ matters: $W=\displaystyle\int_0^LqE_0e^{-\lambda x}dx$.

Step 1. Integrate. $W=\dfrac{qE_0}{\lambda}\left[-e^{-\lambda x}\right]_0^L=\dfrac{qE_0}{\lambda}\left(1-e^{-\lambda L}\right)$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: 314

Concept. Torque on a coil $\tau=NIAB\sin\theta$, $\theta$ between the axis (moment) and $\vec B$.

Step 1. Value. $125\times1\times\pi(0.02)^2\times0.4\times\tfrac12=0.0314$ N m.

Answer. $314$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) connect resistor of $\dfrac{x}{2}\,\Omega$, in series with voltmeter.

Concept. A voltmeter's full-scale current is fixed: $I=\dfrac{20}{x}$. For $30$ V the total resistance must be $\dfrac{30}{I}=1.5x$.

Step 1. Extra. $1.5x-x=\dfrac x2$ in series.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2.64$

Concept. A flat spiral has turns of every radius between $r_1$ and $r_2$; turns per unit radius $\dfrac{N}{r_2-r_1}$. Add $I\pi r^2$ over all turns.

Step 1. Integral. $M=\dfrac{NI\pi}{r_2-r_1}\displaystyle\int_{r_1}^{r_2}r^2dr=\dfrac{NI\pi(r_2^3-r_1^3)}{3(r_2-r_1)}$.

Step 2. Value. $\dfrac{200\times0.02\times\pi\times189\times10^{-6}}{0.09}=2.64\times10^{-2}$ A m$^2$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{2+\pi}{1+\pi}$

Use: a semi-infinite straight wire ending at the foot of the perpendicular from the point gives $\dfrac{\mu_0I}{4\pi r}$; a semicircle gives $\dfrac{\mu_0I}{4r}$ at its centre; a straight wire whose line passes through the point gives zero.

Figure (I): two semi-infinite wires and the semicircle. All three fields at P point the same way (into or out of the page together):

$$B_1=2\cdot\frac{\mu_0I}{4\pi r}+\frac{\mu_0I}{4r}=\frac{\mu_0I}{4\pi r}(2+\pi)$$

Figure (II): the horizontal semi-infinite wire and the semicircle contribute; the vertical wire lies along a line through Q, so it contributes nothing:

$$B_2=\frac{\mu_0I}{4\pi r}+\frac{\mu_0I}{4r}=\frac{\mu_0I}{4\pi r}(1+\pi)$$

$$\frac{B_1}{B_2}=\frac{2+\pi}{1+\pi}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 171

Concept. Magnetic force $\vec F=q\,\vec v\times\vec B$. Work out the cross product with the determinant.

Step 1. Cross product. $(\hat i-2\hat j+3\hat k)\times(2\hat i+3\hat j-5\hat k)=(10-9)\hat i+(6+5)\hat j+(3+4)\hat k=\hat i+11\hat j+7\hat k$.

Step 2. Magnitude. $|\vec F|=10^{-6}\sqrt{1+121+49}=\sqrt{171}\times10^{-6}$ N.

Answer. $\alpha=171$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $50\hat i+100\hat j$

Concept. Lorentz force $\vec F=q(\vec E+\vec v\times\vec B)$. With $\vec v=v_x\hat i+v_y\hat j$ and $\vec B=B\hat k$: $\vec v\times\vec B=v_yB\,\hat i-v_xB\,\hat j$.

Step 1. Components. $x$: $qv_yB=4\times10^{-10}$; $y$: $qE-qv_xB=2\times10^{-10}$.

Step 2. Solve. $qB=4\times10^{-12}$: $v_y=100$; $4\times10^{-10}-4\times10^{-12}v_x=2\times10^{-10}\Rightarrow v_x=50$.

Answer. $50\hat i+100\hat j$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) D Only

Concept. Ampere's law with a circle of radius $r$ inside a uniformly carrying wire encloses $I\dfrac{r^2}{a^2}$, so $B=\dfrac{\mu_0Ir}{2\pi a^2}$: zero on the axis, rising linearly to a maximum at the surface.

Step 1. Check each statement. A: the field does not vary along the length of a long wire: false. B: the axis has the minimum (zero), not the maximum: false. C: the surface has the maximum: false. D: minimum (zero) at the axis: true. E: $B$ changes with $r$: false.

Answer. D only (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) 2

Concept. On a loop's axis $B=B_0\dfrac{R^3}{(R^2+x^2)^{3/2}}$.

Step 1. At $x=\sqrt3R$. $(4R^2)^{3/2}=8R^3$, so $B=\dfrac{16}{8}=2\ \mu$T.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $6\times10^{-6}$ N towards $R$

Concept. Force per length between parallel wires $\dfrac{\mu_0I_1I_2}{2\pi d}$: like currents attract, unlike repel.

Step 1. From P. P (up) and Q (down) are unlike: Q is pushed away from P, towards R. $\dfrac FL=\dfrac{2\times10^{-7}\times3\times1}{0.03}=2\times10^{-5}$ N/m.

Step 2. From R. Q and R both down: Q is pulled towards R. $\dfrac FL=\dfrac{2\times10^{-7}\times2\times1}{0.02}=2\times10^{-5}$ N/m.

Step 3. On 15 cm. $(4\times10^{-5})(0.15)=6\times10^{-6}$ N, towards R.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) 25

Concept. Ammeter: shunt $S$ in parallel carries the extra current; $I_gG=(I-I_g)S$.

Step 1. Solve. $S=\dfrac{1\times100}{5-1}=25\ \Omega$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{3\mu_oI}{4\sqrt2r}$ towards $Q$

Concept. On the axis of a loop, $B=\dfrac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}$. A current that looks clockwise to an observer gives a field pointing away from that observer.

Step 1. One loop at $x=r$. $B=\dfrac{\mu_0Ir^2}{2(2r^2)^{3/2}}=\dfrac{\mu_0I}{4\sqrt2\,r}$.

Step 2. Directions. Seen from O, both currents are clockwise, so P's field at O points towards P and Q's points towards Q: opposite.

Step 3. Net. $\dfrac{\mu_0(4I-I)}{4\sqrt2\,r}=\dfrac{3\mu_0I}{4\sqrt2\,r}$, towards Q (the larger current).

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. Undeflected in crossed fields: $qE=qvB$, so $E=vB$. With $E$ off, circular motion: $r=\dfrac{mv}{qB}$ gives $B$.

Step 1. Field. $B=\dfrac{mv}{qr}=\dfrac{1.6\times10^{-27}\times2\times10^5}{1.6\times10^{-19}\times0.02}=0.1$ T.

Step 2. Electric field. $E=vB=2\times10^4$ N/C.

Answer. $x=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: 1

Concept. A long wire gives $B=\dfrac{\mu_0I}{2\pi r}$, direction by the right-hand rule. Find each field at P with its direction, then add.

Step 1. Distances. P is $4$ cm from Y and $6+4=10$ cm from X.

Step 2. Sizes. $B_X=\dfrac{2\times10^{-7}\times5}{0.10}=1\times10^{-5}$ T; $B_Y=\dfrac{2\times10^{-7}\times4}{0.04}=2\times10^{-5}$ T.

Step 3. Directions. X carries current up and P is to its right: field into the page. Y carries current down: at P its field is out of the page. They oppose.

Step 4. Net. $2\times10^{-5}-1\times10^{-5}=1\times10^{-5}$ T.

Answer. $x=1$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) A, B only

Concept. $S_I=\dfrac{\theta}{I}=\dfrac{NAB}{k}$ and $S_V=\dfrac{S_I}{R}=\dfrac{NAB}{kR}$. Adding turns of the same wire raises $R$ in proportion to $N$.

Step 1. A. $k$ is torque per radian: $[ML^2T^{-2}]$. True.

Step 2. B. Raising $S_I$ by adding turns raises $R$ too, so $S_V$ need not rise. True.

Step 3. C. Doubling $N$ doubles $S_I$ and $R$: $S_V$ is unchanged. False.

Step 4. D. An ammeter needs a shunt of small resistance. False.

Step 5. E. $S_I\propto N$, not $\dfrac1N$. False.

Answer. A, B only (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $149$

Concept. Same shunt formula.

Step 1. Solve. $S=\dfrac{0.02\times30}{3-0.02}=\dfrac{30}{149}\ \Omega$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 8

Each side is at distance $d = a/2$ from the centre and subtends $45^\circ$ on each side:
$$B_{side} = \frac{\mu_0I}{4\pi d}(\sin45^\circ + \sin45^\circ) = \frac{\mu_0I\sqrt2}{4\pi(a/2)}$$

Four sides:
$$B = 4\cdot\frac{\sqrt2\,\mu_0I}{2\pi a} = \frac{2\sqrt2\,\mu_0I}{\pi a} = \frac{2\sqrt2\times4\pi\times10^{-7}\times5}{\pi\times\frac{1}{\sqrt2}} = 2\sqrt2\cdot\sqrt2\times4\times10^{-7}\times5 = 8\times10^{-6}\ \text{T}$$

$p = 8$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) see figure

Concept. Ampere's law: inside, $B=\dfrac{\mu_0Ir}{2\pi a^2}\propto r$; outside, $B=\dfrac{\mu_0I}{2\pi r}\propto\dfrac1r$. The two meet at $r=a$, so the graph is continuous.

Step 1. Shape. A straight line from the origin up to a peak at $r=a$, then a $1/r$ fall.

Step 2. Reject the others. (A) is flat inside; (B) is zero inside and jumps at $r=a$; (C) is largest on the axis.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{N}{2\pi}q\omega$

Concept. The current through a loop is the charge crossing its surface per second. The rotating ring carries $N$ charges past any point $\dfrac{\omega}{2\pi}$ times a second each.

Step 1. Loop A. A wraps round the ring once, so every charge crosses it once per revolution: $I_A=\dfrac{Nq\omega}{2\pi}$.

Step 2. Loop B. The ring passes through B's surface twice, in opposite directions, so the net charge crossing is zero: $I_B=0$.

Step 3. Difference. $I_A-I_B=\dfrac{N}{2\pi}q\omega$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Both (A) and (R) are true and (R) is the correct explanation of (A)

Concept. $\vec F=q\vec v\times\vec B$ is zero when $\vec v\parallel\vec B$.

Step 1. Assertion. An electron moving along the field feels no force, so it can move in a straight line at constant velocity in a uniform field: true.

Step 2. Reason. If the field is along the velocity, $\vec v\times\vec B=0$: true, and this is exactly why the velocity stays constant.

Answer. Both true and R explains A (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 250

Concept. Inside the solenoid $B=\mu_0nI$; the electron circles with $T=\dfrac{2\pi m}{eB}$.

Step 1. Field. $B=\dfrac{2\pi m}{eT}=\dfrac{2\pi\times9\times10^{-31}}{1.6\times10^{-19}\times75\times10^{-9}}=4.71\times10^{-4}$ T.

Step 2. Turns per metre. $n=\dfrac{B}{\mu_0I}=\dfrac{4.71\times10^{-4}}{4\pi\times10^{-7}\times1.5}=250$.

Answer. $250$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $a\,e^{-\frac{4\pi Mv_0}{q\mu_0I}}$

Concept. The magnetic force does no work, so the speed stays $v_0$. Its component along the wire changes: $m\dfrac{dv_z}{dt}=qB\dfrac{dr}{dt}$ with $B=\dfrac{\mu_0I}{2\pi r}$, so $m\,\Delta v_z=\dfrac{q\mu_0I}{2\pi}\ln\dfrac ra$.

Step 1. Turning point. Radial velocity zero; all of $v_0$ is along the wire, now reversed: $v_z=-v_0$.

Step 2. Equation. $m(-2v_0)=\dfrac{q\mu_0I}{2\pi}\ln\dfrac xa$.

Step 3. Solve. $x=a\,e^{-4\pi Mv_0/(q\mu_0I)}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{\mu_0}{4\pi}\dfrac{I}{a}\left[\dfrac{3\pi}{2} + 1\right]$

Concept. Split the wire: the arc gives $\dfrac{\mu_0I\theta}{4\pi a}$; a semi-infinite wire ending at the foot of the perpendicular from O gives $\dfrac{\mu_0I}{4\pi a}$; a wire whose line passes through O gives nothing.

Step 1. Lower straight wire. It comes from infinity along the line $a$ below O and ends directly below O: $\dfrac{\mu_0I}{4\pi a}$. Current to the left, O above it: into the page.

Step 2. Arc. Three-quarters of a circle, $\theta=\dfrac{3\pi}2$, traversed clockwise: $\dfrac{\mu_0I}{4\pi a}\cdot\dfrac{3\pi}2$, into the page.

Step 3. Upper wire. Leaves along the line through O: zero.

Step 4. Add. $B=\dfrac{\mu_0}{4\pi}\dfrac Ia\left[\dfrac{3\pi}2+1\right]$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $[a/2,\ 2a]$

Concept. Ampère's law for a uniform wire: inside $B=\dfrac{\mu_0Ir}{2\pi a^2}$ (grows with $r$); outside $B=\dfrac{\mu_0I}{2\pi r}$. The maximum is at the surface.

Step 1. Inside. Half the maximum at $r=\dfrac a2$.

Step 2. Outside. Half the maximum at $r=2a$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 8

Concept. $B=\mu_0\dfrac NLI\Rightarrow L=\dfrac{\mu_0NI}{B}$. The radius is not needed.

Step 1. Length. $\dfrac{4\pi\times10^{-7}\times200\times0.29}{2.9\times10^{-4}}=0.08\pi$ m.

Answer. $8\pi$ cm.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $19950\ \Omega$

Concept. Voltmeter: series resistance $R=\dfrac{V}{I_g}-G$.

Step 1. Solve. $\dfrac{100}{0.005}-50=19950\ \Omega$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 72

Concept. Same finite-wire formula; a hexagon has six sides, each subtending $30^\circ$ on either side of the perpendicular from the centre.

Step 1. Geometry. Side $\dfrac{4\pi}{6}=\dfrac{2\pi}{3}$ m; distance $d=\dfrac{\sqrt3}{2}\times\dfrac{2\pi}{3}=\dfrac{\pi}{\sqrt3}$ m.

Step 2. One side. $10^{-7}\times\dfrac{4\pi\sqrt3}{\pi/\sqrt3}\times(2\sin30^\circ)=12\times10^{-7}$ T.

Step 3. Six sides. $72\times10^{-7}$ T.

Answer. $72$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{G}{20}$

Concept. Current divides inversely as resistance: $0.05I$ through $G$, $0.95I$ through $S$, so $S=\dfrac{G}{19}$. The ammeter's resistance is $G\parallel S$.

Step 1. Combine. $\dfrac{G\cdot G/19}{G+G/19}=\dfrac{G}{20}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: 4

Concept. In a moving coil galvanometer, magnetic torque balances the spring: $NIAB=k\theta$.

Step 1. Solve. $k=\dfrac{100\times10^{-2}\times2\times10^{-4}\times0.01}{0.05}=4\times10^{-5}$ N m rad$^{-1}$.

Answer. $4$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) the electron will continue to move with uniform velocity along the axis of the solenoid.

The field inside the solenoid is along its axis, parallel to the electron's velocity. So $\vec F = -e(\vec v\times\vec B) = 0$ and the electron keeps moving with uniform velocity along the axis.

Solution by Sreeraj P, M.Sc Physics

Answer: 50

Concept. $B$ depends only on $x$. The forces on the two sides parallel to $x$ are equal and opposite. The two sides parallel to $y$ carry opposite currents in different fields, so the net force is $F=IL\,[B(x_2)-B(x_1)]$.

Step 1. Fields. $x_1=2$ m, $x_2=2.5$ m: $B(x_2)-B(x_1)=0.2\times2\times0.5=0.2$ T.

Step 2. Force. $F=0.5\times0.5\times0.2=0.05$ N.

Answer. $50$ mN.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. Outside both wires, wires with currents in the same direction give fields in the same direction, so the magnitudes add.

Step 1. At A. A is $r$ from the wire with $I$ and $3r$ from the wire with $2I$: $B_A=\dfrac{\mu_0}{2\pi}\left(\dfrac Ir+\dfrac{2I}{3r}\right)=\dfrac{\mu_0I}{2\pi r}\cdot\dfrac53$.

Step 2. At C. C is $r$ from $2I$ and $3r$ from $I$: $B_C=\dfrac{\mu_0}{2\pi}\left(\dfrac{2I}{r}+\dfrac{I}{3r}\right)=\dfrac{\mu_0I}{2\pi r}\cdot\dfrac73$.

Step 3. Ratio. $\dfrac{B_A}{B_C}=\dfrac57$.

Answer. $x=5$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) outside the cable

Concept. Ampère's law outside the cable: the loop encloses $I-I=0$, so $B=0$ there.

Answer. Outside the cable (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. With only the resistance box $R$ in series, $I=\dfrac{E}{R+G}$ and $\theta=\dfrac IK$, where $K$ is the figure of merit. So $\dfrac1\theta=\dfrac{K}{E}(R+G)$: a straight line with slope $\dfrac KE$.

Step 1. Slope. From $(2,1)$ to $(6,2)$: $\dfrac{2-1}{6-2}=0.25$ per ohm.

Step 2. Figure of merit. $K=0.25\times E=0.25\times2=0.5$ A/division.

Step 3. Check. The intercept $0.5=\dfrac{KG}{E}$ gives $G=2\ \Omega$, consistent.

Answer. $5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. The wire floats when $BIL=mg$. Find $L$ from the resistance.

Step 1. Length. $L=\dfrac{RA}{\rho}=\dfrac{1\times10^{-5}}{2\times10^{-6}}=5$ m.

Step 2. Field. $B=\dfrac{0.5\times10}{2\times5}=0.5$ T.

Answer. $5$ ($\times10^{-1}$ T).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $20$

Concept. The voltmeter data give $I_g$; then the shunt formula.

Step 1. Full-scale current. $I_g=\dfrac{10}{500}=0.02$ A.

Step 2. Shunt. $S=\dfrac{0.02\times100}{10-0.02}=0.2\ \Omega$.

Answer. $20\times10^{-2}\ \Omega$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\vec F_1 = q\vec E,\ \vec F_2 = q(\vec v\times\vec B)$

The Lorentz force is $\vec F = q\vec E + q(\vec v\times\vec B)$: electric part $q\vec E$, magnetic part $q(\vec v\times\vec B)$.

Solution by Sreeraj P, M.Sc Physics

Answer: 500

$B = \mu_0\dfrac{m}{l}I$:
$$m = \frac{Bl}{\mu_0I} = \frac{6.28\times10^{-3}\times0.5}{4\pi\times10^{-7}\times5} = 500$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4\times10^{-8}\ \text{T}$

Concept. Biot–Savart for a short element: $dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$; on the $y$-axis, $\theta=90^\circ$.

Step 1. Value. $10^{-7}\times\dfrac{10\times0.01}{0.25}=4\times10^{-8}$ T.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. Work to turn a magnetic moment from $\theta_1$ to $\theta_2$: $W=MB(\cos\theta_1-\cos\theta_2)$.

Step 1. Moment. $M=200\times10^{-4}\times2.5\times10^{-4}=5\times10^{-6}$ A m$^2$.

Step 2. Work ($0\to90^\circ$). $MB=5\times10^{-6}$ J.

Answer. $5\ \mu$J.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Concept. Recall the shape of each dependence.

Step 1. A. For a paramagnetic or diamagnetic material, susceptibility does not depend on the magnetising field: a horizontal line, III.

Step 2. B. Inside a wire, $B\propto x$: a straight line through the origin, I.

Step 3. C. Outside a wire, $B\propto\dfrac1x$: IV.

Step 4. D. Inside a solenoid, $B=\mu_0nI$ is uniform across the section and falls only near the ends: II.

Answer. (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: 160

Concept. The ammeter is a $240\ \Omega$ coil in parallel with a $10\ \Omega$ shunt: one resistor of $\dfrac{240\times10}{250}=9.6\ \Omega$ in the circuit. It reads the total current.

Step 1. Circuit. $R=140.4+9.6=150\ \Omega$.

Step 2. Current. $I=\dfrac{24}{150}=0.16$ A.

Answer. $160$ mA.

Solution by Sreeraj P, M.Sc Physics

Answer: 100

Concept. Plane perpendicular to $\vec B$ means the moment is along $\vec B$ ($\theta=0$). Turning by $90^\circ$ needs $W=MB$.

Step 1. Value. $100\times10^{-3}\times5\times10^{-3}\times0.2=10^{-4}$ J.

Answer. $100\ \mu$J.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) (A), (B) and (D) only

Concept. The velocity stays the same only if the net force $q(\vec E+\vec v\times\vec B)$ is zero.

Step 1. Each case. (A) No fields: no force. Possible. (B) $E=0$, $B\ne0$: zero force if $\vec v\parallel\vec B$. Possible. (C) $E\ne0$, $B=0$: the force $q\vec E$ cannot be cancelled. Not possible. (D) $E\ne0$, $B\ne0$: crossed fields with $E=vB$ cancel. Possible.

Answer. (A), (B) and (D) only (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 160

Each wire is $2.5\ \text{cm}$ from P. For opposite currents, the fields at a point between the wires point in the same direction, so they add:
$$B = 2\times\frac{\mu_0 I}{2\pi d} = 2\times\frac{2\times10^{-7}\times10}{0.025} = 2\times8\times10^{-5} = 1.6\times10^{-4}\ \text{T} = 160\ \mu\text{T}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $60\ \mu\text{A}$

Concept. $k\theta=NIAB$, so $\theta\propto I$.

Step 1. Ratio. $60^\circ=\dfrac\pi3$ rad: $I=200\times\dfrac{\pi/10}{\pi/3}=200\times\dfrac3{10}=60\ \mu$A.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

The straight segments lie along lines through O, so they give no field at O. The current goes round the loop in one sense, so the two semicircles give fields in the same direction:
$$B = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2} = \frac{4\pi\times10^{-7}\times4}{4}\left(\frac{1}{2\pi} + \frac{1}{4\pi}\right)$$
$$B = 4\pi\times10^{-7}\times\frac{3}{4\pi} = 3\times10^{-7}\ \text{T} \;\Rightarrow\; \alpha = 3$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $3.75\times10^{-3}\ \Omega$

$$S = \frac{I_gG}{I - I_g} = \frac{3\times10^{-3}\times10}{8 - 0.003} \approx \frac{0.03}{8} = 3.75\times10^{-3}\ \Omega$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $96\ \Omega$

Concept. Deflection $\propto$ current through the coil. With a shunt, the coil gets $\dfrac{S}{S+G}$ of the current.

Step 1. Fraction. $\dfrac{5}{25}=\dfrac{24}{24+G}$.

Step 2. Solve. $G=96\ \Omega$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\left(\dfrac{R_1}{R_2}\right)^2$

Concept. After acceleration through $V$: $mv=\sqrt{2mqV}$. Radius $r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mqV}}{qB}\propto\sqrt m$ (same $q$, $V$, $B$).

Step 1. Ratio. $\dfrac{m_X}{m_Y}=\left(\dfrac{R_1}{R_2}\right)^2$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 32

$$\vec F = q\vec v\times\vec B = 4\times10^{-6}\times4\times10^6\,(\hat j\times2\hat k) = 16\times2\,\hat i = 32\,\hat i\ \text{N}$$
So $x = 32$.

Solution by Sreeraj P, M.Sc Physics

Answer: 4

$$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2\times5\times1.6\times10^{-19}}{9\times10^{-31}}} = \frac43\times10^6\ \text{m s}^{-1}$$
For no deflection $eE = evB$:
$$E = vB = \frac43\times10^6\times3\times10^{-6} = 4\ \text{N C}^{-1}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1:1$

Concept. Same formulas: $B\left(\tfrac a2\right)=\tfrac12B_{\max}$ and $B(2a)=\tfrac12B_{\max}$.

Answer. $1:1$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.20\ \Omega$

A shunt is connected in parallel:
$$S = \frac{I_gG}{I - I_g} = \frac{20\times10^{-6}\times200}{0.02 - 0.00002} \approx \frac{4\times10^{-3}}{0.02} = 0.20\ \Omega$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Statement I is true but Statement II is false

Concept. Test each statement against the theory.

Step 1. Statement I. With time-varying currents the forces between two circuits need not be equal and opposite at each instant; momentum is conserved only if the momentum carried by the electromagnetic field is counted. True.

Step 2. Statement II. For steady currents Ampere's circuital law can be derived from the Biot–Savart law; the two are equivalent. So it does depend on it: false.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 160

Concept. In a non-uniform field, forces on opposite sides do not cancel. Sides along $x$ cancel each other; the sides along $y$ at $x_1$ and $x_2$ give a net force $IL[B(x_2)-B(x_1)]$.

Step 1. Field difference. $B_0\times4\times2=40$ T.

Step 2. Force. $2\times2\times40=160$ N.

Answer. $160$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\sqrt2:1$

$r = \dfrac{\sqrt{2mK}}{qB}$. Same $q$ and $K$, so $r \propto \sqrt m$:
$$\frac{r_d}{r_p} = \sqrt{\frac{2}{1}} = \sqrt2 : 1$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{\mu_0 I}{\sqrt2\,a}$

Concept. Each loop gives $\dfrac{\mu_0I}{2a}$ along its own axis; perpendicular planes give perpendicular fields.

Step 1. Resultant. $\sqrt2\times\dfrac{\mu_0I}{2a}=\dfrac{\mu_0I}{\sqrt2a}$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 35

Concept. Only the horizontal component (pointing north) acts on a horizontal wire. SE to NW makes $45^\circ$ with north.

Step 1. Force per length. $IB\sin45^\circ=\sqrt2\times3.5\times10^{-5}\times\dfrac1{\sqrt2}=3.5\times10^{-5}$ N/m.

Answer. $35$.

Solution by Sreeraj P, M.Sc Physics

Answer: 40

Concept. Finite wire: $B=\dfrac{\mu_0I}{4\pi d}(\sin\alpha_1+\sin\alpha_2)$. At a square's centre $d=\dfrac a2$ and each angle is $45^\circ$; four sides add.

Step 1. One side. $10^{-7}\times\dfrac{5}{0.5}\times\sqrt2=\sqrt2\times10^{-6}$ T.

Step 2. Four sides. $4\sqrt2\times10^{-6}=40\sqrt2\times10^{-7}$ T.

Answer. $X=40$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $-2iBR\,\hat j$

Concept. In a uniform field the force on any wire equals the force on the straight line joining its ends. The part in the field runs from the left section's exit point, up and round, to the right section's exit point: its ends are $2R$ apart along $x$.

Step 1. Equivalent wire. Length $2R$, current from left to right: $\vec L=2R\,\hat i$.

Step 2. Force. $\vec F=i\vec L\times\vec B=i(2R)B_0(\hat i\times\hat k)=-2iB_0R\,\hat j$.

Answer. $-2iBR\,\hat j$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2\times10^{-4}\ \text{N m}$ along negative $Z$-direction

Concept. Anticlockwise as seen looking along $-x$ means the moment points along $-x$ (right-hand rule). $\vec\tau=\vec m\times\vec B$.

Step 1. Moment. $5\times0.02=0.1$ A m$^2$ along $-\hat i$.

Step 2. Torque. $0.1\times2\times10^{-3}\,(-\hat i\times\hat j)=-2\times10^{-4}\hat k$ N m.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 20

Concept. Coil at its centre: $B=\dfrac{\mu_0NI}{2R}$. Perpendicular coils: add as vectors.

Step 1. Coil P. $\dfrac{4\pi\times10^{-7}\times100\times1}{2\times0.01\pi}=2$ mT. Coil Q. $4$ mT.

Step 2. Resultant. $\sqrt{4+16}=\sqrt{20}$ mT.

Answer. $x=20$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) see figure

Concept. Use symmetry. Reflecting the loop in its own plane ($z\to-z$) leaves the loop and its current unchanged; under this reflection the field components in the plane ($B_x$, $B_y$) change sign. So $B_y(0,a,-z)=-B_y(0,a,z)$: $B_y$ is an odd function of $z$.

Step 1. In the plane. At $z=0$ the field of the loop is along $z$ (perpendicular to the plane), so $B_y=0$.

Step 2. Far away. $B_y\to0$ as $|z|\to\infty$.

Step 3. Shape. Zero at the centre and at both ends, opposite signs on the two sides, with one hump each side: only graph (C) is odd.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{\mu_0I}{\pi a}$

Concept. O lies on the line of AB and DE, so those parts give nothing. BC and ET are semi-infinite wires starting level with O, each at distance $\dfrac a2$: each gives $\dfrac{\mu_0I}{4\pi(a/2)}=\dfrac{\mu_0I}{2\pi a}$.

Step 1. BC. Current down, O to its right: $(-\hat y)\times\hat x=+\hat z$ (out of the page).

Step 2. ET. Current up, O to its left: $\hat y\times(-\hat x)=+\hat z$ (out of the page).

Step 3. Add. $B=2\times\dfrac{\mu_0I}{2\pi a}=\dfrac{\mu_0I}{\pi a}$.

Answer. $\dfrac{\mu_0I}{\pi a}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2\ \text{T}$

Concept. Same wire rewound: $N\times2\pi r$ is fixed, so $r\propto\dfrac1N$ and $B=\dfrac{\mu_0NI}{2r}\propto N^2$.

Step 1. Ratio. From $4$ turns to $1$: $\dfrac{1}{16}$.

Step 2. Value. $\dfrac{32}{16}=2$ T.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $10\ \text{V}$

Concept. Only the top side and the upper parts of the two vertical sides are in the field. The forces on the vertical parts are equal and opposite, so the net force is $BIL$ on the top side, which must hold up the mass.

Step 1. Current. $B=10^3$ G $=0.1$ T: $I=\dfrac{mg}{BL}=\dfrac{10^{-3}\times10}{0.1\times0.1}=1$ A.

Step 2. Voltage. $V=IR=1\times10=10$ V.

Answer. $10$ V (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $N_AI_A=4N_BI_B$

Concept. $M=NI\pi r^2$. Equal moments: $N_AI_Ar_A^2=N_BI_Br_B^2$.

Step 1. Radii squared. $100$ and $400$.

Answer. $N_AI_A=4N_BI_B$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $3\sqrt3\times10^{-5}\ \text{T}$

Concept. Each side of a regular polygon is a finite wire. For an equilateral triangle the centroid is at $d=\dfrac{a}{2\sqrt3}$ from each side, and each end subtends $60^\circ$ with the perpendicular.

Step 1. Distance. $d=\dfrac{4\sqrt3}{2\sqrt3}=2$ cm.

Step 2. One side. $B_1=\dfrac{\mu_0I}{4\pi d}(2\sin60^\circ)=10^{-7}\times\dfrac{2}{0.02}\times\sqrt3=\sqrt3\times10^{-5}$ T.

Step 3. Three sides. All three fields point the same way (the current circulates one way round O): $B=3\sqrt3\times10^{-5}$ T.

Answer. $3\sqrt3\times10^{-5}$ T (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1:4$

Concept. The current enters at the middle of PR and leaves at the middle of QS. The two halves of the loop are identical, so PQ and RS each carry half the current, in the same direction: they attract with $\dfrac FL=\dfrac{\mu_0(I/2)^2}{2\pi d}\propto I^2$.

Step 1. Ratio. $I\to2I$ multiplies the force per length by 4.

Answer. $1:4$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $0\%$

Concept. Current sensitivity $\dfrac{\theta}{I}=\dfrac{NAB}{k}$; voltage sensitivity $\dfrac{\theta}{V}=\dfrac{NAB}{kR}$. More turns of the same wire raise $R$ in the same ratio.

Step 1. Effect. $N\times1.5$ and $R\times1.5$: voltage sensitivity unchanged.

Answer. $0\%$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $N^2:n^2$

Concept. Same as above: $B\propto N^2$ for a fixed wire length.

Answer. $N^2:n^2$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $8F_1$

Concept. Force per length between parallel wires $\dfrac{F}{L}=\dfrac{\mu_0I_1I_2}{2\pi d}$.

Step 1. Factor. Currents doubled: $\times4$; distance halved: $\times2$.

Answer. $8F_1$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2\sqrt2:1$

$B_{centre}=\dfrac{\mu_0i}{2R}$, $B_{axis}=\dfrac{\mu_0iR^2}{2(R^2+x^2)^{3/2}}$. With $x=R$:
$$\frac{B_{centre}}{B_{axis}}=\frac{(2R^2)^{3/2}}{R^3}=2\sqrt2$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $176\times10^{-4}\ \text{T}$

Concept. Long solenoid: $B=\mu_0nI$, with $n$ in turns per metre.

Step 1. Turns. $70$ per cm $=7000$ per m.

Step 2. Field. $4\pi\times10^{-7}\times7000\times2=176\times10^{-4}$ T.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 9

Concept. $F=ILB\sin\theta$, with $\theta$ the angle between the side and $\vec B$ (along the hypotenuse).

Step 1. Angle. In the $5$-$12$-$13$ triangle, the $5$ cm side makes an angle with $\cos\theta=\dfrac{5}{13}$, so $\sin\theta=\dfrac{12}{13}$.

Step 2. Force. $2\times0.05\times0.75\times\dfrac{12}{13}=\dfrac{0.9}{13}=\dfrac{9}{130}$ N.

Answer. $x=9$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) A-III, B-I, C-IV, D-II

Concept. Add the pieces at O with their signs: circle $\dfrac{\mu_0I}{2r}$, semicircle $\dfrac{\mu_0I}{4r}$, infinite wire at distance $r$: $\dfrac{\mu_0I}{2\pi r}$, semi-infinite wire ending at the foot of the perpendicular: $\dfrac{\mu_0I}{4\pi r}$, a wire on a line through O: 0.

Step 1. A (straight wire with a full loop). The loop circulates against the straight wire's field: $\dfrac{\mu_0I}{2r}-\dfrac{\mu_0I}{2\pi r}=\dfrac{\mu_0I}{2\pi r}[\pi-1]$, which is III.

Step 2. B (hairpin). Two semi-infinite wires and a semicircle, all the same way: $\dfrac{\mu_0I}{4\pi r}\cdot2+\dfrac{\mu_0I}{4r}=\dfrac{\mu_0I}{4\pi r}[\pi+2]$, which is I.

Step 3. C. One semi-infinite wire ending below O, a semicircle, and a wire on a line through O: $\dfrac{\mu_0I}{4\pi r}[\pi+1]$, which is IV.

Step 4. D (semicircular dip). The straight parts lie on the line through O: only the semicircle, $\dfrac{\mu_0I}{4r}$, which is II.

Answer. A-III, B-I, C-IV, D-II (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) 1.0

$NIAB=k\theta$:
$$A=\frac{k\theta}{NIB}=\frac{4\times10^{-5}\times0.05}{200\times0.01\times0.01}=\frac{2\times10^{-6}}{0.02}=10^{-4}\ \text{m}^2=1\ \text{cm}^2$$

Solution by Sreeraj P, M.Sc Physics

Answer: 68

Concept. P sees the two wires along perpendicular lines, so the two fields (each perpendicular to its own line) are perpendicular too.

Step 1. Distance. $\dfrac{7}{\sqrt2}$ cm $=0.05$ m (with $\sqrt2=1.4$).

Step 2. Fields. $\dfrac{2\times10^{-7}\times8}{0.05}=32\ \mu$T; $\dfrac{2\times10^{-7}\times15}{0.05}=60\ \mu$T.

Step 3. Resultant. $\sqrt{32^2+60^2}=68\ \mu$T.

Answer. $68$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) B and E only

Concept. $\vec F=q\,\vec v\times\vec B$ with $q=-e$ for an electron. A force perpendicular to $\vec v$ in a uniform field gives a circle.

Step 1. Direction. $\hat i\times(-\hat k)=\hat j$, so for $+e$ the force would be along $+y$; for the electron it is along $-y$.

Step 2. Path. $\vec v\perp\vec B$, constant speed, force always perpendicular: circular motion.

Step 3. Statements. B and E are true; A, C and D are false.

Answer. B and E only (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. The tension in the leads is removed when the magnetic force on AB is upward and equal to the weight: $BIL=mg$.

Step 1. Solve. $I=\dfrac{mg}{BL}=\dfrac{0.04\times10}{0.40\times0.50}=2$ A (flowing so that $I\vec L\times\vec B$ points up).

Answer. $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) (A) and (C)

Concept. Voltmeter: series $R=\dfrac{V}{I_g}-G$. Ammeter: shunt $r=\dfrac{I_gG}{I-I_g}$.

Step 1. Voltmeter. $R=\dfrac{50}{10^{-3}}-54=49\,946\ \Omega\approx50$ k$\Omega$: (A).

Step 2. Ammeter. $r=\dfrac{1\times54}{10-1}=6\ \Omega$: (C).

Answer. (A) and (C) (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 40

Concept. The orbiting electron is a current loop: $I=\dfrac{ev}{2\pi r}$, and $B=\dfrac{\mu_0I}{2r}=\dfrac{\mu_0ev}{4\pi r^2}$.

Step 1. Value. $10^{-7}\times\dfrac{1.6\times10^{-19}\times6.76\times10^6}{(0.52\times10^{-10})^2}=\dfrac{1.08\times10^{-19}}{2.70\times10^{-21}}=40$ T.

Answer. $40$.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. The straight parts lie on the line through O, so they give no field there. Only the semicircle counts: $B=\dfrac{\mu_0I}{4R}$.

Step 1. Value. $B=\dfrac{4\pi\times10^{-7}\times14}{4\times0.022}=\dfrac{\pi\times14\times10^{-7}}{0.022}=2.0\times10^{-4}$ T.

Answer. $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $+25\%$

Voltage sensitivity $=\dfrac{\text{current sensitivity}}{R}$. With $R$ unchanged it rises by the same $25\%$.

Solution by Sreeraj P, M.Sc Physics