First slide

Magnetism

Question

When a short bar magnet is kept in $TanA$ position on a deflection magnetometer, the magnetic needle Oscillates with a frequency $' f'$ and the deflection produced is $45^{0}$ . If the bar magnet is removed the frequency of oscillation of that needle is

Moderate

A

$n_{2} = f (2^{\frac{1}{4}})$
B

$n_{2} = \frac{f}{2^{\frac{1}{4}}}$
C

$n_{2} = f 2^{\frac{1}{3}}$
D

$n_{2} = \frac{f}{2^{\frac{1}{3}}}$

Solution

$W e k n o w, f r e q u e n c y = n = \frac{1}{2 π} \sqrt{\frac{M B}{I}}$
$\begin{array}{l} ∴ n \propto \sqrt{B} \\ \Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{\frac{B_{1}}{B_{2}}} \end{array}$
$I n i t i a l l y i n p r e s e n c e o f b a r m a g n e t, B_{1} = \sqrt{B^{2} + B_{H}^{2}}$
$\begin{array}{l} \Rightarrow B_{1} = \sqrt{(B_{H} \tan 45 °)^{2} + B_{H}^{2}} \\ \Rightarrow B_{1} = \sqrt{2} B_{H} \end{array}$
$\begin{array}{l} F i n a l l y, a f t e r r e m o v i n g b a r m a g n e t, B_{2} = B_{H} \\ ∴ \frac{n_{1}}{n_{2}} = \sqrt{(\frac{\sqrt{2} B_{H}}{B_{H}})} = 2^{\frac{1}{4}} \end{array}$
$\Rightarrow n_{2} = \frac{n_{1}}{2^{\frac{1}{4}}} = \frac{f}{2^{\frac{1}{4}}}$

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