Question

In place a horizontal component of earth's field is $B_H$. The time period of vibration of a magnet is $T$. If we take it to a place where earth's horizontal component is $B_H/2$, then the time period will be :

• A. $T$
• B. $\frac{T}{2}$
• C. $T\sqrt{2}$
• D. $2T$

Answer 1

The correct option is C $T\sqrt{2}$

Time period of vibration of magnet ; T = $2\pi \sqrt{\frac{I}{{mB}_H}}$ ... (1)

I $\Rightarrow$ moment of inertia of magnet about axis of oscillation.

m $\Rightarrow$ magnetic moment.

$B_H\Rightarrow$ horizontal component of earth's magnetic freed.

&1 $B_H$ becomes $\frac{B_H}{2}$ ; T becomes $T^1$

$T^1=2\pi \sqrt{\frac{I}{{mB}_{H/2}}}=2\pi \sqrt{\frac{2.2}{{mB}_{H/2}}}=T\sqrt{2}$

$\Rightarrow T^1=T\sqrt{2}$

OPTION : C.