Question

An infinite straight conductor is kept along \( X'X \) axis and carries a current \( I \). A charge \( q \) at point \( P(0, r) \) starts moving with velocity \( \vec{v} = v_0 \, \hat{j} \) as shown in figure. Find the direction and magnitude of force initially experienced by the charge. 

Hint:

The magnetic force on a moving charge is perpendicular to both the velocity and the magnetic field, as determined by the right-hand rule.

Answer 1

The magnetic field \( \vec{B} \) generated by an infinite straight conductor carrying current \( I \) at distance \( r \) is \( \vec{B} = \frac{\mu_0 I}{2 \pi r} \, \hat{\phi} \), where \( \hat{\phi} \) indicates the azimuthal direction around the wire. The force \( \vec{F} \) on a charge \( q \) moving with velocity \( \vec{v} \) in magnetic field \( \vec{B} \) is given by \( \vec{F} = q (\vec{v} \times \vec{B}) \). Substituting \( \vec{v} = v_0 \, \hat{j} \) and \( \vec{B} = \frac{\mu_0 I}{2 \pi r} \, \hat{\phi} \), we get \( \vec{F} = q \, v_0 \, \hat{j} \times \frac{\mu_0 I}{2 \pi r} \, \hat{\phi} \). Applying the right-hand rule to \( \hat{j} \times \hat{\phi} \) yields \( \hat{i} \) (pointing towards the \( X \)-axis). Therefore, the force is \( \vec{F} = \frac{\mu_0 I q v_0}{2 \pi r} \, \hat{i} \). Final Answer: The force acts towards the \( X \)-axis with a magnitude of \( F = \frac{\mu_0 I q v_0}{2 \pi r} \).