Question

A bar magnet is at right angles to a uniform magnetic field. The couple acting on the magnet is to be one fourth by rotating it from the position. The angle of rotation is

• A. $Si{n}^{-1}\left(0.25\right)$
• B. ${90}^0-Si{n}^{-1}\left(0.25\right)$
• C. $Co{s}^{-1}\left(0.25\right)$
• D. ${90}^0-Co{s}^{-1}\left(0.25\right)$

Answer 1

Answer: (B)

${90}^0-Si{n}^{-1}\left(0.25\right)$

$\overrightarrow{z}=\overrightarrow{m}\times \overrightarrow{B}$
$z_1=mB$
$z_2=\frac{z_1}{4}$
$=\frac{mB}{4}$
$\frac{mB}{4}=mBsin\theta$
$\theta =si{n}^{-1}0.25$
$\therefore Theanglerotated={90}^0-si{n}^{-1}0.25$