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# A magnetic needle of magnetic moment $4.8 \times 10^{-2}\;JT^{-1}$ is placed at $30^{\circ}$ with the direction of uniform magnetic field of magnitude $3 \times 10^{-2} \;T$ Calculate the torque acting on the needle

$\begin{array}{1 1} 7.2 \times 10^{-4}J \\ 7.2 \times 10^{-4}JT^{-1} \\ 7.2 \times 10^{4}J \\ 7.2 \times 10^{4}JT^{-1} \end{array}$

Given : $B= 3 \times 10^{-2}\;T$
$\theta= 30^{\circ}$
$M= 4.8 \times 10^{-2}$
Torque $T= MB \sin \theta$
$\quad= 4.8 \times 10^{-2}$
$\quad= 4.8 \times 10^{-2} \times 3 \times 10^{-2} \times \sin 30^{\circ}$
$\quad= 4.8 \times 10^{-2} \times 3 \times 10^{-2} \times \large\frac{1}{2}$
$\quad= 7.2 \times 10^{-4}J$
Answer : $7.2 \times 10^{-4}J$