Question

A short bar magnet placed in a uniform magnetic field experiences a maximum couple of $4\times {10}^{-5}$N-m. The magnet is turned through ${45}^o$. The couple acting in the new equilibrium position is

• A. $2\times {10}^{-5}N$
• B. $2.828\times {10}^{-5}Nm$
• C. $1.414\times {10}^{-5}Nm$
• D. $3.6\times {10}^{-5}Nm$

Answer 1

The correct option is B $2.828\times {10}^{-5}Nm$

$z_{max}=4\times {10}^{-5}Nm$
$\theta =45$
$z=mB\sin \theta$
$z_{max}=mB$
$z_1=mB\sin {45}^o$
$z_1=\frac{mB}{\sqrt{2}}$
$=\frac{z_{max}}{\sqrt{2}}$
$=\frac{4\times {10}^{-5}}{\sqrt{2}}$
$=2\sqrt{2}\times {10}^{-5}$
$=2.828\times {10}^{-5}Nm$