A solenoid of 0.4111 length with 500 turns carries a current of 3A. A coil of 10 turns and of radius 0.01m carries a current of 0.4A. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is

Options

A. `6pi^(2)xx10^(-7)Nm`
B. `3pi^(2)xx10^(-7)Nm`
C. `9pi^(2)xx10^(-7)Nm`
D. `12pi^(2)xx10^(-7)Nm`

Answer

To solve the problem step by step, we will calculate the magnetic field produced by the solenoid, the magnetic moment of the coil, and finally the torque acting on the coil when placed in the magnetic field of the solenoid.

### Step 1: Calculate the magnetic field (B) inside the solenoid

The formula for the magnetic field inside a solenoid is given by:

\[ B = \mu_0 \cdot \frac{N}{L} \cdot I \]

Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( N = 500 \) (number of turns of the solenoid)
- \( L = 0.4111 \, \text{m} \) (length of the solenoid)
- \( I = 3 \, \text{A} \) (current through the solenoid)

Substituting the values:

\[ B = 4\pi \times 10^{-7} \cdot \frac{500}{0.4111} \cdot 3 \]

Calculating \( \frac{500}{0.4111} \):

\[ \frac{500}{0.4111} \approx 1216.5 \]

Now substituting back into the equation for B:

\[ B \approx 4\pi \times 10^{-7} \cdot 1216.5 \cdot 3 \]

Calculating \( B \):

\[ B \approx 4\pi \times 10^{-7} \cdot 3649.5 \approx 4.58 \times 10^{-3} \, \text{T} \]

### Step 2: Calculate the magnetic moment (m) of the coil

The magnetic moment \( m \) of the coil is given by:

\[ m = n \cdot I \cdot A \]

Where:
- \( n = 10 \) (number of turns of the coil)
- \( I = 0.4 \, \text{A} \) (current through the coil)
- \( A = \pi r^2 \) (area of the coil)

Calculating the area \( A \):

\[ A = \pi \cdot (0.01)^2 = \pi \cdot 0.0001 = 3.14 \times 10^{-4} \, \text{m}^2 \]

Now substituting the values into the equation for m:

\[ m = 10 \cdot 0.4 \cdot 3.14 \times 10^{-4} \]

Calculating \( m \):

\[ m = 4 \cdot 3.14 \times 10^{-4} = 1.256 \times 10^{-3} \, \text{A m}^2 \]

### Step 3: Calculate the torque (τ) on the coil

The torque \( \tau \) on the coil in the magnetic field is given by:

\[ \tau = m \cdot B \cdot \sin(\theta) \]

Since the axis of the coil is at right angles to that of the solenoid, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \):

\[ \tau = m \cdot B \]

Substituting the values we calculated:

\[ \tau = (1.256 \times 10^{-3}) \cdot (4.58 \times 10^{-3}) \]

Calculating \( \tau \):

\[ \tau \approx 5.76 \times 10^{-6} \, \text{N m} \]

### Step 4: Convert to the required format

To express this in terms of \( \pi \):

\[ \tau \approx 6 \pi^2 \times 10^{-7} \, \text{N m} \]

Thus, the torque required to hold the coil is approximately:

\[ \tau \approx 6 \pi^2 \times 10^{-7} \, \text{N m} \]

### Final Answer

The torque required to hold the coil is \( 6 \pi^2 \times 10^{-7} \, \text{N m} \).

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