A magnetic dipole is acted upon by two magnetic fields with 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

Options

A. 1
B. 11
C. 36
D. 1060

Explanation

To solve the problem, we need to analyze the magnetic dipole in the presence of two magnetic fields. Here’s the step-by-step solution:

### Step 1: Understand the Problem
We have two magnetic fields, \( B_1 \) and \( B_2 \), inclined at an angle of \( 75^\circ \) to each other. The dipole is in stable equilibrium at an angle of \( 30^\circ \) with \( B_1 \), which has a magnitude of \( 15 \, \text{mT} \).

### Step 2: Identify Angles
Since the two magnetic fields are inclined at \( 75^\circ \), and the dipole makes an angle of \( 30^\circ \) with \( B_1 \), we can find the angle \( \theta_2 \) that the dipole makes with \( B_2 \):
\[ \theta_2 = 75^\circ - 30^\circ = 45^\circ \]

### Step 3: Write the Torque Equation
For the dipole to be in stable equilibrium, the net torque acting on it must be zero. The torque (\( \tau \)) due to a magnetic field is given by:
\[ \tau = m B \sin \theta \]
where \( m \) is the magnetic moment, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the magnetic moment and the magnetic field.

### Step 4: Set Up the Torque Balance Equation
Since the torques due to both fields must be equal for equilibrium, we can write:
\[ m B_1 \sin(30^\circ) = m B_2 \sin(45^\circ) \]
We can cancel \( m \) from both sides since it is common:
\[ B_1 \sin(30^\circ) = B_2 \sin(45^\circ) \]

### Step 5: Substitute Known Values
Substituting \( B_1 = 15 \, \text{mT} \), \( \sin(30^\circ) = \frac{1}{2} \), and \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ 15 \cdot \frac{1}{2} = B_2 \cdot \frac{1}{\sqrt{2}} \]

### Step 6: Solve for \( B_2 \)
Now, we can solve for \( B_2 \):
\[ \frac{15}{2} = \frac{B_2}{\sqrt{2}} \]
Multiplying both sides by \( \sqrt{2} \):
\[ B_2 = \frac{15 \sqrt{2}}{2} \]

### Step 7: Calculate the Numerical Value
Calculating \( B_2 \):
\[ B_2 = \frac{15 \cdot 1.414}{2} \approx \frac{21.21}{2} \approx 10.605 \, \text{mT} \]
Rounding this gives approximately \( 11 \, \text{mT} \).

### Final Answer
The magnitude of the other field \( B_2 \) is approximately \( 11 \, \text{mT} \).

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