Question 41: Magnetism and Matter

Question

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

Accepted Answer

$$	an 60^\circ  = {{{B_0}\sin \delta } \over {{B_0}\cos \delta \cos 45^\circ }}$$
$$ \Rightarrow 	an \delta  = \sqrt {{3 \over 2}} $$
$$ \Rightarrow \delta  = {	an ^{ - 1}}\left( {\sqrt {{3 \over 2}} } ight)$$

Answer

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

Answer

$$	an ^{-1}(\sqrt{6})$$

Answer

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

Answer

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

Magnetism and Matter

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

Choose the correct answer: