Question

A closely wound solenoid of $2000$ turns and area of cross-section $1.5\times {10}^{-4}{m}^2$ carries a current of $2.0A$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5\times {10}^{-2}$ Tesla making an angle of ${30}^o$ with the axis of the solenoid. The torque on the solenoid will be-

• A. $1.5\times {10}^{-3}N.m$
• B. $1.5\times {10}^{-2}N.m$
• C. $3\times {10}^{-2}N.m$
• D. $3\times {10}^{-3}N.m$

Answer 1

The correct option is B $1.5\times {10}^{-2}N.m$

Given :

Solution :by simply using formula we get,

$\overrightarrow{\tau }=\overrightarrow{M}$ x $\overrightarrow{B}$

$\overrightarrow{\tau }=niA$ x 5x 10 ${}^{-}$ ${}^2$ $\sin \left({30}^o\right)$

$\overrightarrow{\tau }$ = 2000 x 2 x 1.5 x 10 ${}^{-}$ ${}^4$ x 5x 10 ${}^{-}$ ${}^2$x $\frac{1}{2}$

$\overrightarrow{\tau }$ = 0.3 x 5 x10 ${}^{-}$ ${}^2$

$\overrightarrow{\tau }$ = 1.5x 10 ${}^{-}$ ${}^2$ $N.m$

hence option B is correct