Question

In an uniform field the magnetic needle completes 10 oscillation in 92seconds. When a small magnet is placed in the magnetic meridian 10 cm due north of needle with north pole towards south completes 15 oscillation in 69 seconds. The magnetic moment of magnet ($B_H=0.3G$) is

• A. $4.5A{m}^2$
• B. $0.45A{m}^2$
• C. $0.75A{m}^2$
• D. $0.225A{m}^2$

Answer 1

The correct option is C $0.75A{m}^2$

Given; $f_1=\frac{10}{92}=0.10869565217(sec{)}^{-1}$

$f_2=\frac{15}{69}=0.21739130435(sec{)}^{-1}$

$B_H=0.3\ast {10}^{-4}T$

To find: magnetic moment of magnet,$M=?$

Solution: As we know that,

$\frac{B_2}{B_1}=\frac{f_1^2}{f_2^2}$

here, $B_2$ is field of magnet and $B_1=B_H$

$==>B_2=B_H\frac{(0.10869565217{)}^2}{(0.21739130435{)}^2}$ $...\left(1\right)$

also, $B_2=\frac{{\mu }_0}{4\pi }\frac{M}{r^2}$ $...\left(2\right)$

equating $\left(1\right),\left(2\right)$we get

$\frac{{\mu }_0}{4\pi }\frac{M}{r^2}=0.3\ast {10}^{-4}\ast 0.24999999998$

$==>{10}^{-7}\frac{M}{(0.1{)}^2}=0.07499999999\ast {10}^{-4}$

$M=0.75A{m}^2$

hence,

The correct opt : C