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(A) In the given case,the magnetic field $H$ produced by the coil and the horizontal component of the Earth's magnetic field $H_0$ are perpendicular t

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) In the given case,the magnetic field $H$ produced by the coil and the horizontal component of the Earth's magnetic field $H_0$ are perpendicular to each other.When a magnetic needle is placed in two mutually perpendicular magnetic fields $H$ and $H_0$,it aligns itself along the resultant magnetic field.From the geometry of the vector addition,we can form a right-angled triangle where the base is $H$ and the perpendicular side is $H_0$.The angle $	heta$ that the magnet makes with $H$ is given by the trigonometric relation:$	an 	heta = \frac{	ext{Perpendicular}}{	ext{Base}} = \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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