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Question: A bar magnet suspended in a field of induction \( 1.5 \times {10^{ - 6}}\,T\; \) has a time period \...

A bar magnet suspended in a field of induction 1.5×10−6 T  1.5 \times {10^{ - 6}}\,T\; has a time period π/5\pi /5 sec. The moment of inertia about the axis of suspension is 1.5×10−4 kgm21.5 \times {10^{ - 4}}\,kg{m^2} . The magnetic moment of the magnet is
A) 10000 Am210000\,A{m^2}
B) 1/16 Am21/16\,A{m^2}
C) 10 Am210\,A{m^2}
D) 1 Am21\,A{m^2}

Explanation

Solution

Hint : A bar magnet when suspended in a magnetic field will oscillate while trying to align itself to the external magnetic field. The time period of oscillation depends directly on the moment of inertia and inversely on the induction of the external magnetic field and its magnetic moment.

Formula used: In this question, we’ll use the following formula:
⇒T=2πIMB\Rightarrow T = 2\pi \sqrt {\dfrac{I}{{MB}}} where TT is the time period of oscillation of the bar magnet, II is the moment of inertia of the bar magnet, MM is the magnetic moment of the bar magnet, BB is the strength of the external magnetic field.

Complete step by step answer
When the bar magnet is placed in an external magnetic field, it will try to align itself with the external magnetic field such that the poles of the bar magnet align themselves with the opposite poles of the external magnetic field.
In doing so, the bar magnet will oscillate about the equilibrium position for a long time before it comes to rest. The time period of this oscillation is calculated as:
⇒T=2πIMB\Rightarrow T = 2\pi \sqrt {\dfrac{I}{{MB}}}
Substituting the values of B=1.5×10−6 T  B = 1.5 \times {10^{ - 6}}\,T\; , T=π/5T = \pi /5 and I=1.5×10−4 kgm2I = 1.5 \times {10^{ - 4}}\,kg{m^2} , we get
⇒π5=2π1.5×10−4 M×1.5×10−6   \Rightarrow \dfrac{\pi }{5} = 2\pi \sqrt {\dfrac{{1.5 \times {{10}^{ - 4}}\,}}{{M \times 1.5 \times {{10}^{ - 6}}\,\;}}}
Taking the square on both sides, we get
⇒π225=4π21.5×10−4 M×1.5×10−6   \Rightarrow \dfrac{{{\pi ^2}}}{{25}} = 4{\pi ^2}\dfrac{{1.5 \times {{10}^{ - 4}}\,}}{{M \times 1.5 \times {{10}^{ - 6}}\,\;}}
Solving for the magnetic field, we get,
⇒M=4×25×1.5×10−4 1.5×10−6   \Rightarrow M = 4 \times 25 \times \dfrac{{1.5 \times {{10}^{ - 4}}\,}}{{1.5 \times {{10}^{ - 6}}\,\;}}
∴M=10000 Am2\therefore M = 10000\,A{m^2}
Correct option is (A).

Note
Here we have assumed that the bar magnet faces no energy losses and can oscillate continuously. However in reality, there are energy losses associated with air resistance as well in the support that the bar magnet is tied to and suspended in the magnetic field. This phenomenon is used in magnetic compasses by explorers to find out the magnetic North pole.