Magnetism and Matter: NEET and JEE questions with solutions

Answer: (B) $1280\pi^2$

Time period: $T = \dfrac{5}{20} = 0.25$ s.

For a magnetic needle oscillating in a field:
$$T = 2\pi\sqrt{\frac{I}{mB}} \;\Rightarrow\; m = \frac{4\pi^2 I}{T^2 B}$$

$$m = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{(0.25)^2 \times 0.049} = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{3.0625 \times 10^{-3}} = 4\pi^2 \times 3.2 \times 10^{-3}$$
$$m = 12.8\pi^2 \times 10^{-3} = 1280\pi^2 \times 10^{-5}\ \text{A m}^2$$

So $x = 1280\pi^2$.

Solution by Sreeraj P, M.Sc Physics

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

Diamagnetic: small negative susceptibility, $-1 \leq \chi < 0$ (II).

Ferromagnetic: very large positive susceptibility, $\chi >> 1$ (III).

Paramagnetic: small positive susceptibility (IV).

Non-magnetic: $\chi = 0$ (I).

A-II, B-III, C-IV, D-I.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{M}{2}$

Pole strength $m$ stays the same: $M = mL$.

After bending at the middle, each arm has length $\dfrac{L}{2}$ and the arms make $60°$. The two ends and the bend form an equilateral triangle, so the distance between the poles is $\dfrac{L}{2}$.

$$M' = m \cdot \frac{L}{2} = \frac{M}{2}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) Zero

Gauss's law for magnetism: $\displaystyle\oint\vec{B}\cdot d\vec{S} = 0$.

Magnetic monopoles do not exist, and magnetic field lines are closed loops. Every line that enters a closed surface also leaves it, so the net flux is zero.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$

$$\mu = \mu_0(1 + \chi) = 4\pi \times 10^{-7} \times 600 = 2.4\pi \times 10^{-4}\ \text{T m A}^{-1}$$

(The value of the magnetising field is not needed.)

Solution by Sreeraj P, M.Sc Physics