First slide

Magnetic dipole in a uniform magnetic field

Question

A magnetic needle oscillates in horizontal plane with a period T at a place where the angle of dip is 60^o. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

Moderate

A

T
B

$\sqrt{2} T$
C

$\frac{T}{\sqrt{\sqrt{3}}}$
D

$\frac{T}{\sqrt{2}}$

Solution

We know that $T = 2 π \sqrt{(\frac{I}{MBcos θ})}$
At magnetic meridian, in the vertical plane
$∴ T^{'} = 2 π \sqrt{(\frac{I}{MB \sin θ})}$
For angle of dip 60^o
$\frac{T'}{T} = \sqrt{\cot θ} = \sqrt{c o t 60} = \sqrt{\frac{1}{\sqrt{3}}} T' = \frac{T}{\sqrt{\sqrt{3}}}$

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