Question

The work done by a normal magnetic field in revolving a charged particle q in a circular path will be

• A. zero
• B. $MB(1-cos\theta )$
• C. $MB$
• D. $-MB$

Answer 1

Answer: (A)

zero

$\overrightarrow{F_{magnetic}}=q(\overrightarrow{v}\times \overrightarrow{B})$
$\therefore \overrightarrow{F_{magnetic}}.\overrightarrow{v}=0$
Hence, there is no power supplied to the charged particle over a cycle.
$\Rightarrow$ work done is zero.