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 \( \vec{B} \) due to an infinite straight conductor carrying a current \( I \) at a distance \( r \) is given by: \[ \vec{B} = \frac{\mu_0 I}{2 \pi r} \, \hat{\phi}, \] where \( \hat{\phi} \) is the direction of the magnetic field (in the azimuthal direction around the wire). The force on the charge \( q \) moving with velocity \( \vec{v} \) in the magnetic field \( \vec{B} \) is given by: \[ \vec{F} = q (\vec{v} \times \vec{B}). \] Substitute \( \vec{v} = v_0 \, \hat{j} \) and \( \vec{B} = \frac{\mu_0 I}{2 \pi r} \, \hat{\phi} \): \[ \vec{F} = q \, v_0 \, \hat{j} \times \frac{\mu_0 I}{2 \pi r} \, \hat{\phi}. \] Using the right-hand rule, the direction of \( \hat{j} \times \hat{\phi} \) is \( \hat{i} \) (towards the \( X \)-axis). Thus: \[ \vec{F} = \frac{\mu_0 I q v_0}{2 \pi r} \, \hat{i}. \] Final Answer: The force is directed towards the \( X \)-axis with magnitude: \[ F = \frac{\mu_0 I q v_0}{2 \pi r}. \]