A magnet is suspended in the magnetic meridia | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c)Let $M_1$ and $M_2$ be the magnetic moments of magnets and $H$ the horizontal component of earth&rsquo;s field.
We have $	au = MH\sin 	heta $. I

Vedclass · Accepted Answer

$5:8$

Vedclass · Answer

(c)Let $M_1$ and $M_2$ be the magnetic moments of magnets and $H$ the horizontal component of earth&rsquo;s field.
We have $	au = MH\sin 	heta $. If $\phi$ is the twist of wire, then $	au = C\phi,$ $C$ being restoring couple per unit twist of wire
$ \Rightarrow \;C\phi= MH\sin 	heta $
Here ${\phi_1} = ({180^o} - {30^o}) = {150^o}$$ = 150 	imes \frac{\pi }{{180}}$ $rad$
${\phi_2} = ({270^o} - {30^o}) = {240^o}$$ = 240 	imes \frac{\pi }{{180}}$$ rad$
So,$C{\phi_1} = {M_1}H\sin 	heta $ (For deflection $	heta = {30^o}$ of $I$ magnet)
$C{\phi_2} = {M_2}H\sin 	heta $ (For deflection $	heta = {30^o}$ of $II$ magnet)
Dividing $\frac{{{\phi_1}}}{{{\phi _2}}} = \frac{{{M_1}}}{{{M_2}}}$
$ \Rightarrow \frac{{{M_1}}}{{{M_2}}} = \frac{{{\phi_1}}}{{{\phi_2}}} = \frac{{150 	imes \left( {\frac{\pi }{{180}}} ight)}}{{240 	imes \left( {\frac{\pi }{{180}}} ight)}} = \frac{{15}}{{24}} = \frac{5}{8}$
$ \Rightarrow {M_1}:{M_2} = 5:8$

Similar Questions

The angle between the magnetic meridian and geographical meridian is called

The lines of forces due to earth's horizontal component of magnetic field are

A dip circle lies initially in the magnetic meridian, it shows an angle of dip $\delta$ at a place. The dip circle is rotated through an angle $\alpha$ in the horizontal plane and then it shows an angle of dip $\delta^{\prime}$. Hence $\frac{\tan \delta^{\prime}}{\tan \delta}$ is

A compass needle will show which one of the following directions at the earth's magnetic pole