Question 7: In a uniform magnetic field of 0.049 T , a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8 \times 10^{-6} \mathrm{~kg} \mathrm{~m}^{2}. If the magnitude of magnetic moment of the needle is \mathrm{x} \times 10^{-5} \mathrm{Am}^{2}; then the value of ‘ x ‘ is :
(1) 5 \pi^{2}
(2) 128 \pi^{2}
(3) 50 \pi^{2}
(4) 1280 \pi^{2}
Answer: Option (4)
Explanation:
A magnetic needle oscillating in a uniform magnetic field performs angular simple harmonic motion.
The time period of oscillation of a magnetic dipole in a uniform magnetic field is given by
T=2\pi\sqrt{\frac{I}{mB}},
where I is the moment of inertia, m is the magnetic moment,
and B is the magnetic field.
The needle makes 20 oscillations in 5 s, so the time period is
T=\frac{5}{20}=0.25\,\mathrm{s}.
Rearranging the formula for magnetic moment,
m=\frac{4\pi^{2}I}{B T^{2}}.
Substituting the given values,
m=\frac{4\pi^{2}\times 9.8\times10^{-6}}{0.049\times(0.25)^{2}}.
m=\frac{4\pi^{2}\times 9.8\times10^{-6}}{0.049\times0.0625}.
m=\frac{4\pi^{2}\times 9.8\times10^{-6}}{3.0625\times10^{-3}}.
m=1280\pi^{2}\times10^{-5}\,\mathrm{A\,m^{2}}.
Thus, x=1280\pi^{2}.
Hence, the correct answer is option (4).