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Question

Physics Question on Moving charges and magnetism

A dipole having dipole moment M is placed in two magnetic fields of strength B1 and B2 respectively. The dipole oscillates 60 times in 20 seconds in the B1B_1 magnetic field and 60 oscillations in 30 seconds in the B2B_2 magnetic field. Then find the B1B2\frac{B_1}{B_2}

A

32\frac{3}{2}

B

23\frac{2}{3}

C

49\frac{4}{9}

D

94\frac{9}{4}

Answer

94\frac{9}{4}

Explanation

Solution

The time period of oscillation of a dipole in a magnetic field is given by:
T=2π(IMB)T = 2π√(\frac{I}{MB})
where I is the moment of inertia of the dipole about its axis of rotation.
For a given dipole, the moment of inertia and the dipole moment are constant. Therefore, the time period of oscillation is directly proportional to the square root of the magnetic field strength.
Let the time period of oscillation in the B1 magnetic field be T1 and in the B2 magnetic field be T2. Then:
T1T2=(B2B1)\frac{T_1}{T_2} = √(\frac{B_2}{B_1})
60 oscillations in 20 seconds in the B1 magnetic field gives:
T1=2060=13secondsT_1 = \frac{20}{60} = \frac{1}{3} seconds
60 oscillations in 30 seconds in the B2 magnetic field gives:
T2=3060=12secondsT_2 = \frac{30}{60} = \frac{1}{2} seconds
Substituting these values in the above equation, we get:
1312=(B2)(B1)\frac{\frac{1}{3} }{\frac{1}{2}} = √\frac{(B_2)}{(B_1)}
23=B2B1\frac{2}{3} = \sqrt{\frac{B_2}{B_1}}
Squaring both sides, we get:
49=B2B1\frac{4}{9} = \frac{B_2}{B_1}
Therefore, B1B2=94 \frac{B_1}{B_2}=\frac{9}{4}.
Answer. D