Magnetic Force On A Moving Charge

Question

A proton moving with a speed of $2 	imes 10^{6}$ m/s enters a uniform magnetic field of $0.5$ T at right angles to the field lines; what is the magnitude of the magnetic force experienced by it?

RankGuru · Accepted Answer

As stated in NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), a charge $q$ moving with velocity $v$ in a magnetic field $B$ experiences the Lorentz force $F = qvB\sin	heta$, where $	heta$ is the angle between velocity and field. This force is always perpendicular to both the velocity and the field, so it changes direction of motion without doing any work on the particle. Here the proton enters at right angles, so $	heta = 90^{\circ}$ and $\sin	heta = 1$, giving maximum force. Substituting the proton charge $q = 1.6 	imes 10^{-19}$ C, $v = 2 	imes 10^{6}$ m/s and $B = 0.5$ T: $F = (1.6 	imes 10^{-19})(2 	imes 10^{6})(0.5) = 1.6 	imes 10^{-13}$ N. The option $1.6 	imes 10^{-19}$ N is wrong because it is merely the electronic charge and ignores $v$ and $B$. The value $0.8 	imes 10^{-13}$ N is half the correct answer, arising from wrongly using $\sin 30^{\circ}$. The value $3.2 	imes 10^{-13}$ N doubles the charge incorrectly. Plausibility check: the units work out as $	ext{C} \cdot 	ext{m/s} \cdot 	ext{T} = 	ext{N}$, and the magnitude ($\sim 10^{-13}$ N) is reasonable for a subatomic particle in a moderate field.

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Magnetic Force On A Moving Charge

Select the correct option:

🔒 Solution Hidden from View

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Solution

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