Question

A magnet of moment $80A{m}^2$ is placed in a uniform magnetic field of induction $1.8\times {10}^{-5}T$. If each pole of the magnet experiences a force of $25\times {10}^{-3}N$, the length of the magnet is

• A. 0.292cm
• B. 5.76cm
• C. 0.362cm
• D. 2.262cm

Answer 1

The correct option is B 5.76cm

Given: force on each pole, $F=25\times {10}^{-3}N$

magnetic moment, $M=80A{m}^2$

magnetic field, $B=1.8\times {10}^{-5}T$

Solution: Let length of magnet be L.

and we know that,

magnetic moment, $M=m\times L$

$==>m=M/L$ $........\left(1\right)$

and force experienced by each pole is,

$F=m\times B$ $.........\left(2\right)$

using equation 1, substitute value of m in equation 2,

we get,

$F=\frac{M\times B}{L}$

$==>L=\frac{M\times B}{F}$

$L=\frac{80\times 1.8\times {10}^{-5}}{25\times {10}^{-3}}$

$L=5.76\times {10}^{-2}$ $m$

$==>L=5.76$ $cm$

hence,

The correct opt: b