ELECTROMAGNETISM & RELATIVITY - Chapter 4, Exercise 2 Solution
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Force on a Moving Charge in a Magnetic Field

PROBLEM
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A charge of +5 uC moves at 2,000 m/s perpendicular to a 0.3 T
magnetic field. Calculate the force on the charge.

SOLUTION
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Using the magnetic force formula for a charge moving perpendicular to
the field:

    F = q v B

Substitute the given values:

    q = 5 x 10^-6 C
    v = 2000 m/s
    B = 0.3 T

    F = (5 x 10^-6) x 2000 x 0.3
    F = (5 x 10^-6) x 600
    F = 3 x 10^-3 N (0.003 N)

ANSWER: The force on the charge is 3 x 10^-3 N, or 0.003 N.

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WHY THIS WORKS AS AN ANSWER

Because the charge's velocity is stated to be perpendicular to the
magnetic field, the full vector formula F = qv x B simplifies cleanly
to the scalar form F = qvB used here, with no angle-dependent factor
needed. As the chapter's own comparison table notes, this magnetic
force acts perpendicular to both the charge's velocity and the
field itself - meaning it changes the DIRECTION of the charge's
motion (curving its path) without ever doing any work on it or
changing its speed, a genuine structural difference from the electric
force covered in earlier chapters, which can speed a charge up or
slow it down directly.