Answer-
Magnitude of one of the magnetic field is given by B1 = 1.2 × 10–2 T
Suppose that the magnitude of the other field is B2
And the angle between the field is given, θ = 60°
We are givenknow that at stable equilibrium, the angle between the dipole and the field B1, θ1 = 15°
Then the angle between the dipole and the field B2 can be written as –
θ2 =θ – θ1 = 45°
now, the torque due to the field B1 = the torque due to the field B2
Therefore, we get –
MB1 sin θ1 = MB2 sin θ2
where , M represents the magnetic moment of the dipole. So, we have –
\[{{B}_{2}}=\frac{M{{B}_{1}}sin~{{\theta }_{1}}}{M~sin~{{\theta }_{2}}}\]
\[=\frac{(1.2\times {{10}^{2}})x\sin {{15}^{\circ }}}{\sin {{45}^{\circ }}}=4.39\times {{10}^{-3}}T\]
Therefore, magnetic field due to the other magnetic field is 4.39 x 10-3 T.