A solenoid of length `0.4m` and having `500` turns of wire carries a current of `3amp`.Calculate the torque requred to hold a coil (having radius `0.02 cm` current `2A` and turns `50`)in the middle of the solenoid with its axis perpendicular to the axis of the solenoid.`(pi^(2)=10)`

Explanation

To solve the problem, we need to calculate the torque required to hold a coil in the middle of a solenoid. The steps are as follows:

### Step 1: Calculate the Magnetic Field Inside the Solenoid
The magnetic field \( B \) inside a solenoid is given by the formula:
\[ B = \mu_0 n I \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space),
- \( n = \frac{N}{L} \) (number of turns per unit length),
- \( I \) is the current through the solenoid.

Given:
- Length of the solenoid \( L = 0.4 \, \text{m} \),
- Number of turns \( N = 500 \),
- Current \( I = 3 \, \text{A} \).

First, calculate \( n \):
\[ n = \frac{500}{0.4} = 1250 \, \text{turns/m} \]

Now, calculate \( B \):
\[ B = (4\pi \times 10^{-7}) \times (1250) \times (3) \]
Using \( \pi^2 \approx 10 \):
\[ B \approx (4 \times 10^{-7} \times 1250 \times 3) \approx (4 \times 3.75 \times 10^{-4}) = 4.5 \times 10^{-4} \, \text{T} \]

### Step 2: Calculate the Magnetic Moment of the Coil
The magnetic moment \( M \) of the coil is given by:
\[ M = N' I' A \]
where:
- \( N' = 50 \) (number of turns in the coil),
- \( I' = 2 \, \text{A} \) (current through the coil),
- \( A = \pi r^2 \) (area of the coil).

Given the radius \( r = 0.02 \, \text{cm} = 0.0002 \, \text{m} \):
\[ A = \pi (0.0002)^2 = \pi \times 4 \times 10^{-8} \approx 1.25664 \times 10^{-7} \, \text{m}^2 \]

Now, calculate \( M \):
\[ M = 50 \times 2 \times (1.25664 \times 10^{-7}) \approx 100 \times 1.25664 \times 10^{-7} \approx 1.25664 \times 10^{-5} \, \text{A m}^2 \]

### Step 3: Calculate the Torque
The torque \( \tau \) is given by:
\[ \tau = M B \sin \theta \]
Since the axis of the coil is perpendicular to the axis of the solenoid, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \):
\[ \tau = M B \]

Substituting the values:
\[ \tau = (1.25664 \times 10^{-5}) \times (4.5 \times 10^{-4}) \approx 5.65488 \times 10^{-9} \, \text{N m} \]

### Final Result
The torque required to hold the coil in the middle of the solenoid is approximately:
\[ \tau \approx 5.65 \times 10^{-9} \, \text{N m} \]