Question

In an uniform field the magnetic needle completes 10 oscillations in 92seconds. When a small magnet is placed in the magnetic meridian 10cm due north of needle with north pole towards south completes 15 oscillations in 69seconds. 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 A $4.5A{m}^2$

Given: $f_1=\frac{10}{92}=0.10869565217s^{-1}$

$f_2=\frac{15}{69}=0.21739130435s^{-1}$

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

$r=10cm=0.1m$

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

Solution: $\frac{B_1}{B_H}=\frac{f_1^2}{f_2^2}$

$==>B_1=B_H\frac{(0.10869565217{)}^2}{(0.21739130435{)}^2}$

$==>B_1=0.3\ast {10}^{-4}\ast 0.24999999998$

$==>B_1=0.07499999999\ast {10}^{-4}$

also, we know that, $B=\frac{{\mu }_0}{4\pi }\frac{M}{r^2}=B_1$

equating above both equations,we get

${10}^{-7}\frac{M}{({10}^{-1}{)}^2}=0.07499999999\ast {10}^{-4}$

$M=0.07499999999\ast 10$

$M=0.7499999999$

$M=0.75A{m}^2$

hence,

The correct opt : C