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Topic 8 of 15

Magnetic Fields & Forces

Magnetism and electricity are two faces of the same force. A magnetic field exerts a force on any moving charge — perpendicular to both the velocity and the field. This seemingly simple rule explains electric motors, mass spectrometers, the aurora borealis, and the confinement of plasma in fusion reactors.

Key Concepts

Magnetic Field and Force on a Moving Charge

A charge

q

moving with velocity

\vec{v}

in magnetic field

\vec{B}

experiences force

\vec{F} = q\vec{v}\times\vec{B}

. Magnitude:

F = |q|vB\sin\theta

, where

\theta

is the angle between

\vec{v}

and

\vec{B}

. The force is always perpendicular to

\vec{v}

, so it changes direction but never speed or kinetic energy.

Circular Motion in a Magnetic Field

When a charged particle moves perpendicular to a uniform

\vec{B}

, the magnetic force provides centripetal acceleration. The orbital radius is

r = mv/(|q|B)

and the period is

T = 2\pi m/(|q|B)

— independent of speed (cyclotron principle).

Magnetic Force on a Current-Carrying Wire

A straight wire of length

L

carrying current

I

in field

\vec{B}

experiences force

F = BIL\sin\theta

. Direction from the right-hand rule (or

\vec{F} = I\vec{L}\times\vec{B}

Torque on a Current Loop

A flat current loop of

N

turns, area

A

, carrying current

I

in field

B

experiences torque

\tau = NIAB\sin\theta

, where

\theta

is the angle between the loop's magnetic dipole moment

\vec{\mu} = NIA\hat{n}

and

\vec{B}

. This is the operating principle of galvanometers and electric motors.

Hall Effect

When a current-carrying conductor is placed in a perpendicular magnetic field, charge carriers are deflected sideways, building up a transverse voltage (the Hall voltage). The sign of the Hall voltage reveals whether carriers are positive or negative.

Key Equations

Magnetic force on a charge

F = |q|vB\sin\theta

Magnitude of force on charge q moving at speed v in field B; θ is the angle between v and B. Zero if parallel.

Radius of circular orbit

r = \frac{mv}{|q|B}

Radius of circular path for a particle of mass m and charge q moving perpendicular to field B.

Force on a current-carrying wire

F = BIL\sin\theta

Force on a straight wire of length L carrying current I in field B; θ between wire and B.

Torque on a current loop

\tau = NIAB\sin\theta

Torque on a loop of N turns, area A, current I in field B; θ between magnetic moment and B.

Worked Example

Proton in a Uniform Magnetic Field

Problem

A proton ( $m = 1.673\times10^{-27}$ kg, $q = 1.602\times10^{-19}$ C) enters a region of uniform $B = 0.50$ T with speed $v = 3.0\times10^6$ m/s perpendicular to the field. Find (a) the radius of its circular orbit and (b) its orbital period.

Solution

(a) The magnetic force provides centripetal force: $|q|vB = mv^2/r$ , so:

r = \frac{mv}{|q|B} = \frac{(1.673\times10^{-27})(3.0\times10^6)}{(1.602\times10^{-19})(0.50)} = \frac{5.02\times10^{-21}}{8.01\times10^{-20}} = 0.0626\text{ m} \approx 6.26\text{ cm}

(b) Period (circumference divided by speed):

T = \frac{2\pi r}{v} = \frac{2\pi(0.0626)}{3.0\times10^6} = 1.31\times10^{-7}\text{ s} \approx 131\text{ ns}

Note: $T = 2\pi m/(|q|B)$ is independent of speed — this is the cyclotron principle.

Answer r ≈ 6.26 cm; T ≈ 131 ns.

Practice

Exercises

7 problems

1 of 7

A charge $q = +2\,\mu$ C moves at $v = 3.0\times10^5$ m/s perpendicular to a field $B = 0.40$ T. What is the magnitude of the magnetic force (in N) on it?

Your answer

2 of 7

A straight wire carries $I = 5.0$ A and has length $L = 0.20$ m in a uniform field $B = 0.80$ T perpendicular to the wire. What is the force (in N) on the wire?

Your answer

3 of 7

A proton ( $m = 1.673\times10^{-27}$ kg, $q = 1.602\times10^{-19}$ C) moves at $v = 2.0\times10^6$ m/s perpendicular to $B = 0.30$ T. What is the radius (in cm) of its circular orbit?

Unlock Exercise 3