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JAMB Physics 2024 Paper

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Question : 84

Total: 125

Calculate the magnetic force on an electron in a magnetic field of flux density 10T, with a velocity of 3 x 10

^{7}

m/s at 60º to the magnetic field (e = 1.6 x 10

^{-19}

C)

4.8 x 10 $^{-11}$ N
4.6 x 10 $^{-12}$ N
4.16 x 10 $^{-11}$ N
4.3 x 10 $^{-11}$ N

Solution:

Concept:

The magnetic force on a moving charge is given by

F = qvB\sin\theta

.

Explanation:

The electron has charge

q = e = 1.6 \times 10^{-19}\,\text{C}

.

Its velocity is

v = 3 \times 10^7\,\text{m/s}

.

The magnetic flux density is

B = 10\,\text{T}

.

The angle to the field is

\theta = 60^\circ

.

Substitute into the formula:

F = (1.6 \times 10^{-19})(3 \times 10^7)(10)\sin 60^\circ

\sin 60^\circ = 0.8660

F = 4.8 \times 10^{-11} \times 0.8660

F \approx 4.16 \times 10^{-11}\,\text{N}

Answer:

F \approx 4.16 \times 10^{-11}\,\text{N}

Correct option: C.

4.16 \times 10^{-11}\,\text{N}

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