Test Index

WAEC Physics 2016 Paper

Question : 41

Total: 47

Solution:

Concept:

A charge moving perpendicular to a magnetic field experiences a magnetic force given by

F = qvB\sin\theta

. Since

\theta = 90^\circ

\sin\theta = 1

Explanation:

Given charge,

q = 10\,\mu\text{C} = 10 \times 10^{-6}\,\text{C}

Given velocity,

v = 1.0 \times 10^{5}\,\text{m/s}

Given magnetic flux density,

B = 5.0 \times 10^{-4}\,\text{T}

The velocity is at right angle to the field, so the force is maximum.

Using

F = qvB

F = (10 \times 10^{-6}) \times (1.0 \times 10^{5}) \times (5.0 \times 10^{-4})

F = (10 \times 1.0 \times 5.0) \times (10^{-6} \times 10^{5} \times 10^{-4})

F = 50 \times 10^{-5}\,\text{N}

F = 5.0 \times 10^{-4}\,\text{N}

This matches option B.

Answer:

The force on the charge is

5.0 \times 10^{-4}\,\text{N}

Correct option: B.

5.0 \times 10^{-4}\,\text{N}

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