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:
The conductors are placed along x-axis in X–Y plane. There is a point P located between the conductors

Updated On: Mar 19, 2025
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The Correct Option is C

Solution and Explanation

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

\(\overrightarrow F=\overrightarrow {qv} ×\overrightarrow B\)

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

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

\(200µ_0\hat j–300µ_0\hat i\)

\(4π×10^{−5}(2\hat j–3\hat i)\)

\(Hence,\)  \(x = 3\)

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