A magnet hung at 45^{circ} with the magnetic | vedclass

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Question

(A) The apparent angle of dip $	heta^{\prime}$ in a plane making an angle $\alpha$ with the magnetic meridian is given by the relation: $	an 	heta^

Vedclass · Accepted Answer

$	an ^{-1}\left(\sqrt{\frac{3}{2}}ight)$

Vedclass · Answer

(A) The apparent angle of dip $	heta^{\prime}$ in a plane making an angle $\alpha$ with the magnetic meridian is given by the relation: $	an 	heta^{\prime} = \frac{	an 	heta}{\cos \alpha}$,where $	heta$ is the actual angle of dip.Given,$	heta^{\prime} = 60^{\circ}$ and $\alpha = 45^{\circ}$.Substituting these values into the formula:$	an 60^{\circ} = \frac{	an 	heta}{\cos 45^{\circ}}$$\sqrt{3} = \frac{	an 	heta}{1/\sqrt{2}}$$	an 	heta = \sqrt{3} 	imes \frac{1}{\sqrt{2}} = \sqrt{\frac{3}{2}}$Therefore,the actual angle of dip is $	heta = 	an ^{-1}\left(\sqrt{\frac{3}{2}}ight)$.

$A$ magnet hung at $45^{\circ}$ with the magnetic meridian makes an angle of $60^{\circ}$ with the horizontal. The actual value of the angle of dip is.

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