A cicular coil of 200 turns and radius `10cm` is placed in a uniform magnetic field of `0*5T`, normal to the plane of the coil. If the current in the coil is `3*0A`, calculate the (a) total torque on the coil (b) total force on the coil (c) average force on each electron in the coil, due to the magnetic field. Assume that the area of cross-section of the wire to be `10^-5m^2` and the free electron density is `10^(29)m^-3`.

Explanation

To solve the problem step by step, we will calculate the total torque on the coil, the total force on the coil, and the average force on each electron in the coil due to the magnetic field.

### Given Data:
- Number of turns in the coil, \( n = 200 \)
- Radius of the coil, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Magnetic field strength, \( B = 0.5 \, \text{T} \)
- Current in the coil, \( I = 3.0 \, \text{A} \)
- Area of cross-section of the wire, \( A_{wire} = 10^{-5} \, \text{m}^2 \)
- Free electron density, \( n_e = 10^{29} \, \text{m}^{-3} \)

### Step 1: Calculate the Area of the Coil
The area \( A \) of the circular coil can be calculated using the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.1)^2 = \pi (0.01) \approx 0.0314 \, \text{m}^2 \]

### Step 2: Calculate the Total Torque on the Coil
The torque \( \tau \) on the coil in a magnetic field is given by:
\[ \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \]
Since the magnetic field is normal to the plane of the coil, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \):
\[ \tau = n \cdot I \cdot A \cdot B \]
Substituting the values:
\[ \tau = 200 \cdot 3.0 \cdot 0.0314 \cdot 0.5 \]
Calculating:
\[ \tau = 200 \cdot 3.0 \cdot 0.0314 \cdot 0.5 = 9.42 \, \text{Nm} \]

### Step 3: Calculate the Total Force on the Coil
In a closed loop, the total force \( F \) on the coil in a uniform magnetic field is zero because the forces on opposite sides of the coil cancel each other out. Thus:
\[ F = 0 \, \text{N} \]

### Step 4: Calculate the Average Force on Each Electron in the Coil
The force on an electron in a magnetic field can be expressed as:
\[ F_e = q \cdot v_d \cdot B \]
Where \( q \) is the charge of an electron (\( q = 1.6 \times 10^{-19} \, \text{C} \)) and \( v_d \) is the drift velocity.

From the current equation:
\[ I = n_e \cdot q \cdot A_{wire} \cdot v_d \]
We can rearrange to find \( v_d \):
\[ v_d = \frac{I}{n_e \cdot q \cdot A_{wire}} \]
Substituting the known values:
\[ v_d = \frac{3.0}{10^{29} \cdot 1.6 \times 10^{-19} \cdot 10^{-5}} \]
Calculating \( v_d \):
\[ v_d = \frac{3.0}{1.6 \times 10^{5}} \approx 1.875 \times 10^{-5} \, \text{m/s} \]

Now substituting \( v_d \) back into the force equation:
\[ F_e = q \cdot v_d \cdot B = 1.6 \times 10^{-19} \cdot 1.875 \times 10^{-5} \cdot 0.5 \]
Calculating:
\[ F_e \approx 1.6 \times 10^{-19} \cdot 9.375 \times 10^{-6} \approx 1.5 \times 10^{-24} \, \text{N} \]

### Summary of Results:
(a) Total Torque on the Coil: \( \tau \approx 9.42 \, \text{Nm} \)
(b) Total Force on the Coil: \( F = 0 \, \text{N} \)
(c) Average Force on Each Electron: \( F_e \approx 1.5 \times 10^{-24} \, \text{N} \)