Question

Time period of qscillations of a magnet of magnetic moment M and moment of inertia I in a vertical plane perpendicular to the magnetic meridan at a place where earth's horizontal and vertical component of mgnetic field are $B_{H}and{B}_V$ respectively is

• A. $T=2\pi {\sqrt{\frac{I}{M(B_v^2+B_H^2)}}}^{1/2}$
• B. $T=2\pi \sqrt{\frac{I}{M{B}_v}}$
• C. $T=2\pi \sqrt{\frac{I}{M{B}_H}}$
• D. NONE

Answer 1

The correct option is C $T=2\pi \sqrt{\frac{I}{M{B}_H}}$

Given: magnetic moment of magnet$=M$

and moment of inertia$=I$

it is in a plane $\perp$ to magnetic meridian.

Also, $B_H=horizontal$ $component$

$B_V=vertical$ $component$

To find: Time period of oscillation,$T=?$

Solution: As we know that plane $passingthrough$ to

magnetic axis is known as magnetic meridian.

and,

Plane $\perp$ to magnetic axis is known as magnetic equator.

So, angle of dip is,$\delta =0^{\circ }$

Let total field is B.

Now, $B_H=Bcos\delta ==>B_H=B$

and, $B_V=Bsin\delta ==>B_V=0$

So, total field is, $B=B_H$ $....\left(1\right)$

now we know that, time period of oscillation is,

$T=2\pi \sqrt{\frac{I}{MB}}$

using $e{q}^n\left(1\right)$,we get

$==>T=2\pi \sqrt{\frac{I}{M{B}_H}}$

hence,

The correct opt : C