First slide

Magnetic dipole in a uniform magnetic field

Question

A magnet makes 40 oscillations per minute at a place having magnetic field intensity H = 0.1 x 10^-5T. At another place, it takes 2.5 sec to complete one vibration. The value of earth's horizontal field at that place

Easy

A

0.25 x 10^-6 T
B

0.36 x 10^-6 T
C

0.66 x 10^-6 T
D

1.2 x 10^-6 T

Solution

Frequency of oscillation
n = 40 rpm = 0.67rps
$∴ Time period = \frac{1}{n} = 1 .5 \sec . Time period T = 2 π \sqrt{(I / MH)} \propto \sqrt{1 / H} or \frac{T_{1}}{T_{2}} = \sqrt{(\frac{H_{2}}{H_{1}})} or H_{2} = H_{1} {(T_{1} / T_{2})}^{2} ∴ H_{2} = 0 \cdot 1 \times 10^{- 5} {(\frac{1 \cdot 5}{2 \cdot 5})}^{2} = 0 \cdot 36 \times 10^{- 6} T$

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