If a solenoid is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field? Magnetic moment is `0.6 JT^-1`

Options

A. 0.075Nm
B. 0.080Nm
C. 0.081Nm
D. 0.091Nm

Explanation

To find the magnitude of the torque on the solenoid when its axis makes an angle of \(30^\circ\) with the direction of the applied magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Magnetic field strength, \(B = 0.25 \, \text{T}\)
- Magnetic moment, \(M = 0.6 \, \text{J/T}\)
- Angle, \(\theta = 30^\circ\)

### Step 2: Write the formula for torque
The torque \(\tau\) acting on a magnetic dipole in a magnetic field is given by the formula:
\[ \tau = M \cdot B \cdot \sin(\theta) \]

### Step 3: Substitute the known values into the formula
Now, substituting the given values into the torque formula:
\[ \tau = 0.6 \, \text{J/T} \cdot 0.25 \, \text{T} \cdot \sin(30^\circ) \]

### Step 4: Calculate \(\sin(30^\circ)\)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 5: Substitute \(\sin(30^\circ)\) into the equation
Now we can substitute \(\sin(30^\circ)\) into the equation:
\[ \tau = 0.6 \cdot 0.25 \cdot \frac{1}{2} \]

### Step 6: Perform the multiplication
Calculating the values:
\[ \tau = 0.6 \cdot 0.25 = 0.15 \]
\[ \tau = 0.15 \cdot \frac{1}{2} = 0.075 \, \text{N m} \]

### Step 7: Final result
Thus, the magnitude of the torque on the solenoid when its axis makes an angle of \(30^\circ\) with the direction of the applied field is:
\[ \tau = 0.075 \, \text{N m} \quad \text{or} \quad 7.5 \times 10^{-2} \, \text{N m} \]

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