JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 10)
A charge Q is moving $$\overrightarrow {dl} $$ distance in the magnetic field $$\overrightarrow {B} $$. Find the value of work done by $$\overrightarrow {B} $$.
Zero
$$-$$1
Infinite
1
Explanation
We know,
$$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$$
$$ \therefore $$ $$\overrightarrow F \bot \overrightarrow v $$ and $$\overrightarrow F \bot \overrightarrow B $$
Power (p) $$ = {{dw} \over {dt}}$$
$$ = {{\overrightarrow F .\,\overrightarrow {dl} } \over {dt}}$$
$$ = \overrightarrow F \,.\,{{\overrightarrow {dl} } \over {dt}}$$
$$ = \overrightarrow F \,.\,\overrightarrow v $$
$$ = Fv\cos \theta $$
$$ = Fv\cos 90^\circ $$
$$ = 0$$
As power supply by the field is zero so, total work done also zero.
$$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$$
$$ \therefore $$ $$\overrightarrow F \bot \overrightarrow v $$ and $$\overrightarrow F \bot \overrightarrow B $$
Power (p) $$ = {{dw} \over {dt}}$$
$$ = {{\overrightarrow F .\,\overrightarrow {dl} } \over {dt}}$$
$$ = \overrightarrow F \,.\,{{\overrightarrow {dl} } \over {dt}}$$
$$ = \overrightarrow F \,.\,\overrightarrow v $$
$$ = Fv\cos \theta $$
$$ = Fv\cos 90^\circ $$
$$ = 0$$
As power supply by the field is zero so, total work done also zero.
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