Motion in electromagnetic fields practice questions.

A proton enters a region of uniform magnetic field with its velocity perpendicular to the field. Which statement correctly describes its subsequent motion?

• It follows a circular path at constant speed
• It follows a circular path with continuously increasing speed
• It follows a parabolic path at constant speed
• It travels in a straight line while slowing down

Answer: It follows a circular path at constant speed

The magnetic force is always perpendicular to the velocity, so it acts as a centripetal force: it changes the direction of motion but does no work, leaving the speed constant.

A charged particle moves in a circle of radius r in a uniform magnetic field. Both the particle's speed and the magnetic flux density are now doubled. The new radius of the circular path is:

• r/2
• 4r
• r
• 2r

Answer: r

The radius is r = mv/(qB), so r is proportional to v/B. Doubling both v and B leaves the ratio, and therefore the radius, unchanged.

A straight wire of length 0.50 m carries a current of 3.0 A at right angles to a uniform magnetic field of flux density 0.20 T. Calculate the magnetic force on the wire.

• 0.030 N
• 0.30 N
• 1.2 N
• 3.0 N

Answer: 0.30 N

F = BIL = 0.20 × 3.0 × 0.50 = 0.30 N. A common error is misplacing the decimal point or forgetting one of the three factors.

In a velocity selector, an electric field of strength 2.0 × 10⁴ V m⁻¹ is crossed with a magnetic field of flux density 0.40 T. Determine the speed of charged particles that pass through undeflected.

• 8.0 × 10³ m s⁻¹
• 2.0 × 10⁻⁵ m s⁻¹
• 8.0 × 10⁴ m s⁻¹
• 5.0 × 10⁴ m s⁻¹

Answer: 5.0 × 10⁴ m s⁻¹

For zero deflection the electric and magnetic forces balance: qE = qvB, so v = E/B = 2.0 × 10⁴ / 0.40 = 5.0 × 10⁴ m s⁻¹. Multiplying E by B instead of dividing is a classic error.

An electron travels horizontally due north through a region where the magnetic field points vertically downward. The direction of the magnetic force on the electron is:

• due east
• due west
• due north
• vertically upward

Answer: due east

For a positive charge moving north in a downward field, F = qv × B points west; because the electron is negative, the force is reversed, pointing east. Forgetting the negative charge is the standard trap.

Two long parallel wires carry currents in the same direction and experience a force per unit length F on each other. Both currents are doubled and the separation of the wires is also doubled. The new force per unit length is:

• F
• 2F
• 4F
• 8F

Answer: 2F

The force per unit length is proportional to I₁I₂/d. Doubling both currents multiplies it by 4, while doubling the separation divides it by 2, giving a net factor of 2.