Question

In a region of a uniform electric field \(\vec{E}\), a negatively charged particle is moving with a constant velocity \(\vec{v} = -v_0 \hat{i}\) near a long straight conductor coinciding with \(XX'\) axis and carrying current \(I\) towards \(-X\) axis. The particle remains at a distance \(d\) from the conductor.
(i) Draw diagram showing direction of electric and magnetic fields.
(ii) What are the various forces acting on the charged particle ?
(iii) Find the value of \(v_0\) in terms of \(E\), \(d\) and \(I\).

Hint:

The speed is constant, so the electric force balances the magnetic force: \(|q|E = |q|v_0B\) with \(B = \mu_0 I/(2\pi d)\).

Answer 1

Step 1: Understanding the Concept:
The particle moves at a constant velocity, so its acceleration is zero and the net force on it is zero. Two fields act on it: the uniform electric field \(\vec{E}\) and the magnetic field of the current-carrying conductor. The forces from these two fields must cancel. Weight is neglected as it is very small for a charged particle.

Step 2: Part (i): Directions of the fields:
Take the conductor along the \(X\)-axis with current \(I\) towards \(-X\). Let the particle be at distance \(d\) above the conductor, on the \(+Y\) side, and let \(Z\) point out of the page.

By the right-hand rule, for a current towards \(-X\) the magnetic field at a point above the wire points into the page, that is along \(-\hat{k}\). Its magnitude is
\[ B = \frac{\mu_0 I}{2\pi d} \]
The particle is negative and moves along \(-X\), which is the direction of the current. The magnetic force \(\vec{F}_B = q\,\vec{v} \times \vec{B}\) works out to point along \(+Y\), away from the wire. (This agrees with the idea that a negative charge moving along \(-X\) acts like a current along \(+X\), opposite to the wire current, so it is repelled.)

For the net force to be zero, the electric force must point along \(-Y\). A negative charge feels a force opposite to \(\vec{E}\), so \(\vec{E}\) points along \(+Y\).

Diagram in words: draw the wire along \(X\) with an arrow towards \(-X\). Mark the particle at height \(d\) with velocity arrow towards \(-X\). Draw \(\vec{E}\) upward (\(+Y\)) and \(\vec{B}\) into the page at the particle. Draw \(\vec{F}_B\) upward and \(\vec{F}_E\) downward on the particle.

Step 3: Part (ii): Forces on the particle:
1. Electric force \(\vec{F}_E = q\vec{E}\), of magnitude \(|q|E\), directed along \(-Y\) (towards the wire).

2. Magnetic force \(\vec{F}_B = q\,\vec{v}\times\vec{B}\), of magnitude \(|q|\,v_0 B\), directed along \(+Y\) (away from the wire).

The two are equal and opposite, so \(\vec{F}_E + \vec{F}_B = 0\).

Step 4: Part (iii): Find the value of v0:
Balance the magnitudes of the two forces:
\[ |q|\,E = |q|\,v_0 B \quad \Rightarrow \quad v_0 = \frac{E}{B} \]
Put \(B = \dfrac{\mu_0 I}{2\pi d}\):
\[ v_0 = \frac{2\pi d\,E}{\mu_0 I} \]

Final Answer:
The speed of the particle is given below. Electric and magnetic forces are equal and opposite on it. \[ \boxed{v_0 = \dfrac{2\pi d E}{\mu_0 I}} \]