Question

A coil of area $100c{m}^2$ having $500$ turns carries a current of $1mA$. It is suspended in a uniform magnetic field of induction ${10}^{-3}Wb/m^2$ . It's plane makes an angle of ${60}^o$ with the lines of induction. The torque acting on the coil is :

• A. $250\times {10}^{-8}Nm$
• B. $25\times {10}^{-8}Nm$
• C. $2.5\times {10}^{-8}Nm$
• D. $0.2\times {10}^{-8}Nm$

Answer 1

The correct option is A $250\times {10}^{-8}Nm$

$J=\overrightarrow{M}\times \overrightarrow{B}$
$=ni\overrightarrow{A}\times \overrightarrow{B}$
$=1\times {10}^{-3}\times 100\times {10}^{-4}\times {10}^{-3}\times 500\times \sin {30}^o$
$=25\times {10}^{-7}Nm.$
Since. Plane makes an angle of ${60}^o$ with the lines of induction, it will makes an angle of ${30}^o$ with it's normal.