First slide

Magnetic dipole in a uniform magnetic field

Question

A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where dip angle is 30^o and 15 oscillations per min. at a place where dip angle is 60^o. The ratio of the total earth's magnetic field at the two Places is

Moderate

A

$3 \sqrt{3} : 8$
B

$16 : 9 \sqrt{3}$
C

4 : 9
D

$2 \sqrt{3} : 9$

Solution

Let the total magnetic fields due to earth at the two places be B₁ and B₂. lf horizontal components be (B_H)₁ and (B_H)₂respectively, then
${(B_{H})}_{1} = B_{1} \cos 30^{\circ} and {(B_{H})}_{2} = B_{2} \cos 60^{\circ} Here T_{1} = 3 sec. and T_{2} = 4 sec. \begin{array}{l} T_{1} = 3 = 2 π \sqrt{(\frac{I}{{MB}_{1} \cos 30^{\circ}})} \\ and T_{2} = 4 = 2 π \sqrt{(\frac{I}{{MB}_{2} \cos 60^{\circ}})} \end{array} \begin{array}{l} ∴ \frac{3}{4} = {(\frac{B_{2} \cos 60^{\circ}}{B_{1} \cos 30^{\circ}})}^{1 / 2} or \frac{B_{1}}{B_{2}} = \frac{16}{9} \times \frac{\cos 60^{\circ}}{\cos 30^{\circ}} \\ or \frac{B_{1}}{D} = \frac{16}{0} \times \frac{1}{2} \times \frac{2}{\sqrt{2}} = \frac{16}{0} or B_{1} : B_{2} = 16 : 9 \sqrt{3} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App