Two long parallel conductors $S_1$ and $S_2$ are separated by a distance $10$ cm and carrying currents of $4$ A and $2$ A respectively. The conductors are placed along $x$-axis in $X$-$Y$ plane. There is a point $P$ located between the conductors (as shown in figure).
A charge particle of $3\pi$ coulomb is passing through the point $P$ with velocity $\vec v = (2\hat i + 3\hat j)\ \text{m s}^{-1}$; where $\hat i$ & $\hat j$ represents unit vector along $x$ & $y$ axis respectively.
The force acting on the charge particle is $4\pi\times10^{-5}(-x\hat i + 2\hat j)$ N. The value of $x$ is
Answer: (C) $3$
$P$ is $4$ cm below $S_1$ and $6$ cm above $S_2$. Both currents are along $+x$.
$S_1$ (above $P$) gives a field into the page ($-\hat k$):
$$B_1 = \frac{\mu_0 I_1}{2\pi r_1} = \frac{2\times10^{-7}\times4}{0.04} = 2\times10^{-5}\ \text{T}$$
$S_2$ (below $P$) gives a field out of the page ($+\hat k$):
$$B_2 = \frac{2\times10^{-7}\times2}{0.06} = \frac23\times10^{-5}\ \text{T}$$
Net: $\vec B = -\dfrac43\times10^{-5}\,\hat k$ T.
$$\vec F = q\vec v\times\vec B = 3\pi(2\hat i + 3\hat j)\times\left(-\frac43\times10^{-5}\hat k\right) = -4\pi\times10^{-5}(-2\hat j + 3\hat i)$$
$$\vec F = 4\pi\times10^{-5}(-3\hat i + 2\hat j)\ \text{N} \Rightarrow x = 3$$
Solution by Sreeraj P, M.Sc Physics