A charge Q is moving $\overrightarrow {dl}$ distance in the magnetic field $\overrightarrow {B}$. Find the value of work done by $\overrightarrow {B}$.
Solution
We know,<br><br>$$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$$<br><br>$\therefore$ $\overrightarrow F \bot \overrightarrow v$ and $\overrightarrow F \bot \overrightarrow B$<br><br>Power (p) $= {{dw} \over {dt}}$<br><br>$= {{\overrightarrow F .\,\overrightarrow {dl} } \over {dt}}$<br><br>$= \overrightarrow F \,.\,{{\overrightarrow {dl} } \over {dt}}$<br><br>$= \overrightarrow F \,.\,\overrightarrow v$<br><br>$= Fv\cos \theta$<br><br>$= Fv\cos 90^\circ$<br><br>$= 0$<br><br>As power supply by the field is zero so, total work done also zero.
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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