Verified Solution • 2025–26 Session moderate

QJEEM21Ph3014: JEE Main 2021 Physics - Class 12 JEEM21Ph3014 Solution: A charge Q is moving distance in the magnetic field . Find the…

A charge Q is moving dl→\overrightarrow {dl} distance in the magnetic field B→\overrightarrow {B} . Find the value of work done by B→\overrightarrow {B} .

Multiple Choice Options

  • A Zero
  • B −- 1
  • C Infinite
  • D 1

Step-by-Step Verified Solution

We know,

F→=q(v→×B→)\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)

∴\therefore F→⊥v→\overrightarrow F \bot \overrightarrow v and F→⊥B→\overrightarrow F \bot \overrightarrow B

Power (p) =dwdt= {{dw} \over {dt}}

=F→. dl→dt= {{\overrightarrow F .\,\overrightarrow {dl} } \over {dt}}

=F→ . dl→dt= \overrightarrow F \,.\,{{\overrightarrow {dl} } \over {dt}}

=F→ . v→= \overrightarrow F \,.\,\overrightarrow v

=Fvcos⁡θ= Fv\cos \theta

=Fvcos⁡90∘= Fv\cos 90^\circ

=0= 0

As power supply by the field is zero so, total work done also zero.

Related Learning Resources & Exam Preparation