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JAMB Physics 2012 Paper

Question : 34

Total: 42

Solution:

Concept:

The magnetic force on a moving charge is given by

F = qvB\sin\theta

Since the particle enters at right angles to the field,

\theta = 90^\circ

and

\sin\theta = 1

Explanation:

Use the formula

F = Bqv

Rearrange to find the magnetic field:

B = \frac{F}{qv}

Substitute the given values:

B = \frac{1.8 \times 10^{-8}\ \text{N}}{(1.0 \times 10^{-8}\ \text{C})(3.0 \times 10^{-2}\ \text{m/s})}

Simplify the denominator:

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

B = 6.0 \times 10^{1}\ \text{T} = 60\ \text{T}

Answer:

The magnitude of the magnetic field is

6.0 \times 10^{1}\ \text{T}

Correct option: B.

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