Question 77: Magnetism and Matter

Question

A small bar magnet placed with its axis at 30o with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is :

Accepted Answer

Torque on a bar magnet :
$$	au  = MB\,\sin 	heta $$
Here, $$	heta $$ = 30ÃÂº, I = 0.018 N-m, B = 0.06 T
$$0.018 = M 	imes 0.06 	imes 0.5$$
$$ \Rightarrow M = 0.6\,A{m^2}$$
$$W = {U_f} - {U_i}$$
$$ = MB(\cos {	heta _i} - \cos {	heta _f})$$
$$ = 0.6 	imes 0.06(1 - ( - 1))$$
$$ = 7.2 	imes {10^{ - 2}}\,J$$

Answer

6.4 $$ 	imes $$ 10-2 J

Answer

9.2 $$ 	imes $$ 10-3 J

Answer

7.2 $$ 	imes $$ 10-2 J

Answer

11.7 $$ 	imes $$ 10-3 J

Magnetism and Matter

A small bar magnet placed with its axis at 30o with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is :

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