If `theta =90,` then the magnitude of magnetic force is ?

Options

A. `F _(m) = qv B`
B. 1
C. Zero
D. `10 ^(-8)`

Explanation

To solve the question regarding the magnitude of magnetic force when \( \theta = 90^\circ \), we can follow these steps:

### Step-by-Step Solution

1. **Understand the Formula**: The magnetic force \( F_m \) acting on a charged particle moving in a magnetic field is given by the formula:
\[ F_m = q v B \sin \theta \]
where:
- \( F_m \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

2. **Substitute the Given Value**: In this problem, it is given that \( \theta = 90^\circ \). We will substitute this value into the formula:
\[ F_m = q v B \sin(90^\circ) \]

3. **Evaluate \( \sin(90^\circ) \)**: The sine of \( 90^\circ \) is a well-known trigonometric value:
\[ \sin(90^\circ) = 1 \]

4. **Simplify the Expression**: Now, substituting \( \sin(90^\circ) = 1 \) into the equation gives us:
\[ F_m = q v B \cdot 1 \]
Thus, we can simplify it to:
\[ F_m = q v B \]

5. **Conclusion**: Therefore, the magnitude of the magnetic force when \( \theta = 90^\circ \) is:
\[ F_m = q v B \]

### Final Answer
The magnitude of the magnetic force is \( F_m = q v B \).

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