A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75°. One of the fields has a magnitude of 15 mT. The dipole attains stable equilibrium at an angle of 30° with this field. The magnitude of the other field (in mT) is close to

Let the magnitudes of the two magnetic fields be denoted by $$B_1$$ and $$B_2$$. We are told

$$B_1 = 15\ \text{mT},\qquad B_2 = ?$$

The angle between the two fields is given as

$$\alpha = 75^\circ.$$

The magnetic dipole finally comes to rest (stable equilibrium) along the direction of the resultant field $$\mathbf R = \mathbf B_1 + \mathbf B_2$$. At equilibrium the dipole therefore makes the same angle with $$\mathbf B_1$$ as the resultant field does. This angle is stated to be

$$\theta = 30^\circ.$$

So, in the triangle formed by the vectors $$\mathbf B_1,\ \mathbf B_2,\ \mathbf R$$, we know the angle between $$\mathbf B_1$$ and $$\mathbf R$$ (namely $$\theta$$) and the angle between $$\mathbf B_1$$ and $$\mathbf B_2$$ (namely $$\alpha$$).

To relate the magnitudes we resolve $$\mathbf B_2$$ into components parallel and perpendicular to $$\mathbf B_1$$. Using elementary vector addition, the magnitude of the perpendicular component of $$\mathbf B_2$$ is

$$B_2 \sin\alpha,$$

while the component of $$\mathbf B_2$$ along $$\mathbf B_1$$ is

$$B_2 \cos\alpha.$$

The resultant $$\mathbf R$$ therefore has

• a component along $$\mathbf B_1$$ equal to $$B_1 + B_2\cos\alpha$$, and

• a component perpendicular to $$\mathbf B_1$$ equal to $$B_2\sin\alpha.$$

The angle $$\theta$$ that $$\mathbf R$$ makes with $$\mathbf B_1$$ is then obtained from simple trigonometry:

$$\tan\theta \;=\;\frac{\text{perpendicular component}}{\text{parallel component}} \;=\;\frac{B_2\sin\alpha}{B_1 + B_2\cos\alpha}.$$

We now substitute the known numerical values:

$$\tan 30^\circ \;=\;\frac{B_2\sin 75^\circ}{15 + B_2\cos 75^\circ}.$$

Next we insert the standard trigonometric numbers

$$\tan 30^\circ = \frac{1}{\sqrt 3} \approx 0.577,\qquad \sin 75^\circ \approx 0.9659,\qquad \cos 75^\circ \approx 0.2588.$$

This gives

$$0.577 =\frac{B_2(0.9659)}{15 + B_2(0.2588)}.$$

We now carry out the algebra step by step.

First cross-multiply:

$$0.577\bigl(15 + 0.2588\,B_2\bigr) = 0.9659\,B_2.$$

Multiply out the brackets on the left:

$$0.577 \times 15 + 0.577 \times 0.2588\,B_2 = 0.9659\,B_2.$$

Calculate the numerical products:

$$8.655 + 0.1493\,B_2 = 0.9659\,B_2.$$

Bring the $$B_2$$ terms together on one side:

$$8.655 = 0.9659\,B_2 - 0.1493\,B_2.$$

Combine the coefficients of $$B_2$$ on the right:

$$8.655 = (0.9659 - 0.1493)\,B_2 = 0.8166\,B_2.$$

Finally, solve for $$B_2$$ by dividing both sides by $$0.8166$$:

$$B_2 = \frac{8.655}{0.8166} \approx 10.6\ \text{mT}.$$

The closest option to $$10.6\ \text{mT}$$ is $$11\ \text{mT}$$.

Hence, the correct answer is Option B.