Question

A magnetic dipole is suspended in a region where two uniform magnetic fields are inclined at \(75^\circ\) with each other. In stable equilibrium, if the dipole makes an angle of \(30^\circ\) with the direction of field of \(10\sqrt2\,\text{mT}\), then the value of the second magnetic field is

• A. \(20\sqrt2\,\text{mT}\)
• B. \(10\sqrt2\,\text{mT}\)
• C. \(10\,\text{mT}\)
• D. \(20\,\text{mT}\)

Hint:

For two magnetic fields making an angle \(\theta\), \[ \boxed{ \tan\alpha= \frac{B_2\sin\theta} {B_1+B_2\cos\theta}, } \] where \(\alpha\) is the angle made by the resultant field with \(B_1\). A freely suspended magnetic dipole always aligns along the resultant magnetic field.

Answer 1

The Correct Option is C

Step 1: Identify the direction of the resultant magnetic field. A magnetic dipole in stable equilibrium aligns along the resultant magnetic field. The two fields are inclined at \[ 75^\circ. \] The dipole makes an angle of \[ 30^\circ \] with the first field. Hence, it makes an angle \[ 75^\circ-30^\circ=45^\circ \] with the second field.

Step 2: Use the direction formula for the resultant field. If the first field is \[ B_1=10\sqrt2\,\text{mT}, \] and the second field is \(B_2\), then \[ \tan30^\circ = \frac{B_2\sin75^\circ} {B_1+B_2\cos75^\circ}. \] Substituting \[ \tan30^\circ=\frac1{\sqrt3}, \] \[ \sin75^\circ=\frac{\sqrt6+\sqrt2}{4}, \] \[ \cos75^\circ=\frac{\sqrt6-\sqrt2}{4}, \] and \[ B_1=10\sqrt2, \] we obtain \[ B_2=10\,\text{mT}. \]

Step 3: State the result. Therefore, \[ \boxed{B_2=10\,\text{mT}.} \] Hence, the correct option is \(\boxed{(C)}\).