Question

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 $3π$ coulomb is passing through the point P with velocity $\overrightarrow v=(2\hat i+3\hat j)$ m/s; where $\hat i$ and $\hat j$ represents unit vector along x & y axis respectively. The force acting on the charge particle is $4π×10^{−5}(−x\hat i+2\hat j)N$. The value of x is:

• A. 2
• B. 1
• C. 3
• D. -3

Answer 1

Answer: (C)

3

Answer 2

Field at P is

=$\bigg(\frac{µ_0×i_1}{2πr_1}–\frac{µ_0i_2}{2πr_2}\bigg)\bigg(−\hat k\bigg)$

=$−\bigg(\frac{µ_04}{2π×0.04}−\frac{µ_0×2}{2π×0.06}\bigg)\hat k=–\frac{µ_0×200}{6π}\hat k$

Therefore, the force

$\overrightarrow F=\overrightarrow {qv} ×\overrightarrow B$

= $3π(2\hat i+3\hat j)×\bigg(−\bigg(\frac{µ_0×200}{6π}\bigg)\hat k\bigg)$

=$3π\bigg(\frac{200µ_0}{3π\hat j}−\frac{100µ_0}{π}\hat i)$

= $200µ_0\hat j–300µ_0\hat i$

= $4π×10^{−5}(2\hat j–3\hat i)$

$Hence,$ $x = 3$