Question 226 of 276

A circular coil having $N$ turns and radius $r$ carries a current $I.$ It is held in the $XZ$ plane in a magnetic field $B\hat{i}$ . The torque on the coil due to the magnetic field is:

A$B\pi {r}^{2}IN$B$\dfrac{B{r}^{2}I}{\pi N}$C$\dfrac{B\pi {r}^{2}I}{N}$D$Zero$

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Answer: (A) $B\pi {r}^{2}IN$

Magnetic moment $\left(\vec{M}\right)=NI\left(\pi {r}^{2}\right)\hat{j}$
Torque $\left(\vec{\tau }\right)=\vec{M}\times \vec{B}$
$=\left(NI\left(\pi {r}^{2}\right)\hat{j}\right)\times (B\hat{i})$
$=BNI\pi {r}^{2}(-\hat{k})Nm$
$\left|\vec{\tau }\right|=BNI\pi {r}^{2}Nm$

JEE Main Physics practice question on Magnetic Effects & Magnetism (Force and Torque on Current Carrying Conductor), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.