JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 18)
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 :
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 :
6.4 $$ \times $$ 10-2 J
9.2 $$ \times $$ 10-3 J
7.2 $$ \times $$ 10-2 J
11.7 $$ \times $$ 10-3 J
Explanation
Torque on a bar magnet :
$$\tau = MB\,\sin \theta $$
Here, $$\theta $$ = 30º, I = 0.018 N-m, B = 0.06 T
$$0.018 = M \times 0.06 \times 0.5$$
$$ \Rightarrow M = 0.6\,A{m^2}$$
$$W = {U_f} - {U_i}$$
$$ = MB(\cos {\theta _i} - \cos {\theta _f})$$
$$ = 0.6 \times 0.06(1 - ( - 1))$$
$$ = 7.2 \times {10^{ - 2}}\,J$$
$$\tau = MB\,\sin \theta $$
Here, $$\theta $$ = 30º, I = 0.018 N-m, B = 0.06 T
$$0.018 = M \times 0.06 \times 0.5$$
$$ \Rightarrow M = 0.6\,A{m^2}$$
$$W = {U_f} - {U_i}$$
$$ = MB(\cos {\theta _i} - \cos {\theta _f})$$
$$ = 0.6 \times 0.06(1 - ( - 1))$$
$$ = 7.2 \times {10^{ - 2}}\,J$$
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