A current-carrying coil is placed with its ax | vedclass

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(A) The magnetic field inside the loop is $H$,which acts along the axis of the coil. The horizontal component of the earth's magnetic field is $H_0$,w

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{H_0}{H}ight)$

Vedclass · Answer

(A) The magnetic field inside the loop is $H$,which acts along the axis of the coil. The horizontal component of the earth's magnetic field is $H_0$,which acts in the $N-S$ direction.Since the axis of the coil is perpendicular to the $N-S$ direction,the two magnetic fields $H$ and $H_0$ are perpendicular to each other.$A$ suspended magnet in this region will align itself with the resultant magnetic field.Let the angle made by the magnet with the field $H$ be $	heta$.From the geometry of the vector addition of the two perpendicular fields,we have:$	an 	heta = \frac{	ext{Perpendicular component}}{	ext{Base component}} = \frac{H_0}{H}$Therefore,$	heta = 	an^{-1}\left(\frac{H_0}{H}ight)$.

$A$ current-carrying coil is placed with its axis perpendicular to the $N-S$ direction. Let the horizontal component of the earth's magnetic field be $H_0$ and the magnetic field inside the loop be $H$. If a magnet is suspended inside the loop,it makes an angle $\theta$ with $H$. Then $\theta =$

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