Question

A particle of charge \( q \) is moving with a velocity \( \vec{v} \) at a distance \( d \) from a long straight wire carrying a current \( I \) as shown in the figure. At this instant, it is subjected to a uniform electric field \( \vec{E} \) such that the particle keeps moving undeviated. In terms of unit vectors \( \hat{i}, \hat{j}, \) and \( \hat{k} \), find:
the magnetic field \( \vec{B} \),
the magnetic force \( \vec{F}_m \), and
the electric field \( \vec{E} \) acting on the charge.

Hint:

To find the direction of the magnetic field around a current-carrying wire, use the right-hand rule: Point your thumb in the direction of the current, and your fingers curl in the direction of \( \vec{B} \).

Answer 1

(a) Magnetic Field B

The magnetic field due to a long straight current-carrying wire is given by Ampère’s law:

B = μ₀I / 2πd

The right-hand rule states that the direction of B at a distance d above the wire is along the positive k-direction (out of the plane). Thus:

B = μ₀I / 2πd

(b) Magnetic Force F_m

The force on a charged particle moving in a magnetic field is given by:

F_m = q(v × B)

Given that:

• v = v̅ i (along the x-axis),
• B = B̅ k (along the z-axis),

Using the cross product:

v × B = (v̅ i) × (B̅ k)

Using the vector identity i × k = −j, we get:

F_m = qvB(−j)

Thus:

F_m = −qvB̅ j

(c) Electric Field E

Since the charge moves undeviated, the net force on the particle must be zero. This means that the electric force F_e = qE must cancel out the magnetic force:

qE = −F_m

Substituting F_m = −qvB̅ j:

Dividing by q:

E = vB̅ j

Quick Tip: The required electric field is:

E = vB̅ j