A ciruclar coil of 20 turns and radius `10cm` is placed in a uniform magnetic field of `0.1T` normal to the plane of the coil . If the current in the coil is `5.0A` what is the average force on each electron in the coil due to the magnetic field `(` The coil is made of copper wire of cross`-` sectional area `10^(-5)m^(2)` and the free electron density in copper is given to be about `10^(29)m^(-3))`.

Options

A. `2.5xx10^(-25)N`
B. `5xx10^(-25)N`
C. `4xx10^(-25)N`
D. `3xx10^(-25)N`

Explanation

To find the average force on each electron in the coil due to the magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Number of turns in the coil (N) = 20
- Radius of the coil (r) = 10 cm = 0.1 m
- Magnetic field (B) = 0.1 T
- Current in the coil (I) = 5.0 A
- Cross-sectional area of the wire (A) = \(10^{-5} m^2\)
- Free electron density in copper (n) = \(10^{29} m^{-3}\)
- Charge of an electron (e) = \(1.6 \times 10^{-19} C\)

### Step 2: Calculate the drift velocity (V) of electrons
The drift velocity (V) can be calculated using the formula:
\[ V = \frac{I}{n \cdot e \cdot A} \]
Where:
- \(I\) = current (5.0 A)
- \(n\) = free electron density (\(10^{29} m^{-3}\))
- \(e\) = charge of an electron (\(1.6 \times 10^{-19} C\))
- \(A\) = cross-sectional area (\(10^{-5} m^2\))

Substituting the values:
\[ V = \frac{5.0}{(10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-5})} \]

### Step 3: Calculate the drift velocity (V)
Calculating the denominator:
\[ n \cdot e \cdot A = (10^{29}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-5}) = 1.6 \times 10^{5} \]
Now substituting back to find V:
\[ V = \frac{5.0}{1.6 \times 10^{5}} = 3.125 \times 10^{-5} m/s \]

### Step 4: Calculate the magnetic force (F) on each electron
The magnetic force on each electron can be calculated using the formula:
\[ F = e \cdot V \cdot B \]
Substituting the values:
\[ F = (1.6 \times 10^{-19}) \cdot (3.125 \times 10^{-5}) \cdot (0.1) \]

### Step 5: Calculate the force (F)
Calculating the force:
\[ F = 1.6 \times 10^{-19} \cdot 3.125 \times 10^{-6} = 5.0 \times 10^{-25} N \]

### Final Answer
The average force on each electron in the coil due to the magnetic field is \(5.0 \times 10^{-25} N\).

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