Question

Lorentz magnetic force is acting on a particle of charge \(q\) moving with velocity \(\overset{⃗}{V}\) in magnetic field \(\overset{⃗}{B}\). The work done by this force on the charged particle is

• A. zero
• B. \(\overset{⃗}{V}\times \overset{⃗}{B}\)
• C. \(\overset{⃗}{V}\times \overset{⃗}{V}\)
• D. \(q(\overset{⃗}{V}\times \overset{⃗}{B})\)

Hint:

The force is perpendicular to the velocity.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Work done by a force is \(W=\vec F\cdot\vec s\). The Lorentz magnetic force is \(\vec F=q(\vec V\times\vec B)\).

Step 2: Direction:
The force is perpendicular to both \(\vec V\) and \(\vec B\). So it is always perpendicular to the velocity, and hence to the displacement in a small time interval.

Step 3: Work:
\[ dW=\vec F\cdot\vec V\,dt=q(\vec V\times\vec B)\cdot\vec V\,dt=0 \]

Step 4: Choose:
The work done is zero, option (A). Options (B), (C) and (D) are vectors, but work is a scalar.

Final Answer:
A magnetic force does no work on a moving charge. \[ \boxed{0} \]