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Step-by-Step Solution
Step 1: Identify the Condition for Equilibrium
When a magnetic dipole is placed in two magnetic fields, the net torque on the dipole must be zero at equilibrium. Let the two magnetic fields be $B_1$ and $B_2$, with magnitudes of $15\,\text{mT}$ and $B_2$ respectively. Let $m$ be the magnetic dipole moment.
For net torque to be zero, the torque produced by one field must balance the torque by the other field:
$\tau_1 = \tau_2$
Step 2: Write the Torque Balancing Equation
Torque on a magnetic dipole in a magnetic field is given by:
$\tau = m B \sin\theta$
where $\theta$ is the angle between the dipole moment and the magnetic field. Here,
$m B_1 \sin \theta_1 = m B_2 \sin \theta_2$
Step 3: Relate the Angles in the Problem
The two magnetic fields are inclined at $75^\circ$ to each other. The dipole is at an angle of $30^\circ$ with respect to the first field of magnitude $B_1 = 15\,\text{mT}$. Therefore, the angle of the dipole with the second field is:
$\theta_2 = 75^\circ - 30^\circ = 45^\circ$
Step 4: Substitute Known Values
From the torque balance,
$B_1 \sin\theta_1 = B_2 \sin\theta_2$
Substituting $\theta_1 = 30^\circ$, $\theta_2 = 45^\circ$, and $B_1 = 15\,\text{mT}$:
$15 \sin 30^\circ = B_2 \sin 45^\circ$
Recall $\sin 30^\circ = \tfrac{1}{2}$ and $\sin 45^\circ = \tfrac{1}{\sqrt{2}}$:
$15 \times \tfrac{1}{2} = B_2 \times \tfrac{1}{\sqrt{2}}$
Step 5: Solve for $B_2$
$\frac{15}{2} = B_2 \frac{1}{\sqrt{2}}$
$B_2 = \frac{15}{2} \times \sqrt{2}$
$B_2 = 7.5 \times 1.414 \approx 10.6\,\text{mT}$
Rounding to the nearest integer, we get
$B_2 \approx 11\,\text{mT}$
Final Answer
Thus, the magnitude of the other magnetic field is closest to $11\,\text{mT}$.
