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# A particle carrying a charge of 1. 0 x 10-8C enters a magnetic field at 3.0x 10-...

### Question

A particle carrying a charge of 1. 0 x 10-8C enters a magnetic field at 3.0x 10-2 ms -1 at right angles to the field. If the force on this particle is 1.8 x 10-8N, what is the magnitude of the field?

### Options

A) 6.0x10-1T

B) 6.0x10T

C) 6.0x10-3T

D) 6.0x10-4T

The correct answer is B.

### Explanation:

F = BQV

B = $$\frac{F}{QV}$$

B = $$\frac{1.8 \times 10^{-8}}{1.0 \times 10^{-8} \times 3.0 \times 10^{-2}}$$

## Dicussion (1)

• F = BQV

B = $$\frac{F}{QV}$$

B = $$\frac{1.8 \times 10^{-8}}{1.0 \times 10^{-8} \times 3.0 \times 10^{-2}}$$