A test charge having a charge of `1.6 xx 10 ^(-19) C` is moving with a velocity of 10 m/s, and the magnetic field is `25Wbm^(-2)` at an angle of 90 degrees. Determine the magnitude of force ?

Options

A. 22 N
B. 44 N
C. 6.5 N
D. `4 xx 10 ^(-17) N`

Explanation

To determine the magnitude of the magnetic force acting on a test charge moving in a magnetic field, we can use the formula:

\[ F_m = q v B \sin \theta \]

where:
- \( F_m \) is the magnetic force,
- \( q \) is the charge,
- \( v \) is the velocity of the charge,
- \( B \) is the magnetic field strength, and
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

### Step-by-step Solution:

1. **Identify the given values:**
- Charge, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Velocity, \( v = 10 \, \text{m/s} \)
- Magnetic field, \( B = 25 \, \text{Wb/m}^2 \)
- Angle, \( \theta = 90^\circ \)

2. **Substitute the values into the formula:**
Since \( \sin 90^\circ = 1 \), we can simplify the equation:
\[ F_m = q v B \sin \theta = q v B \cdot 1 = q v B \]

3. **Calculate the magnetic force:**
\[ F_m = (1.6 \times 10^{-19} \, \text{C}) \times (10 \, \text{m/s}) \times (25 \, \text{Wb/m}^2) \]
\[ F_m = 1.6 \times 10^{-19} \times 10 \times 25 \]
\[ F_m = 1.6 \times 10^{-19} \times 250 \]
\[ F_m = 400 \times 10^{-19} \, \text{N} \]

4. **Convert to scientific notation:**
\[ F_m = 4.0 \times 10^{-17} \, \text{N} \]

### Final Answer:
The magnitude of the force is \( 4.0 \times 10^{-17} \, \text{N} \).