12.2 Magnetism and Moving Charges

1. Example 1Calculator allowed

   A proton in a magnetic field

   A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m/s perpendicular to a uniform 0.50 T magnetic field. Find the magnetic force on it, the radius of its circular path and the period of its motion.

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   1. 1Step 1: Force: $F = qvB = (1.60 \times 10^{-19})(2.0 \times 10^6)(0.50) = 1.6 \times 10^{-13}\text{ N}$.
   2. 2Step 2: Radius: $r = \dfrac{mv}{qB} = \dfrac{(1.67 \times 10^{-27})(2.0 \times 10^6)}{(1.60 \times 10^{-19})(0.50)} \approx 0.042\text{ m}$.
   3. 3Step 3: Period: $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(1.67 \times 10^{-27})}{(1.60 \times 10^{-19})(0.50)} \approx 1.3 \times 10^{-7}\text{ s}$.
   4. 4Step 4: The magnetic force does no work, so the proton keeps its speed the whole time.

   Answer: F = 1.6 × 10⁻¹³ N, r ≈ 4.2 cm, T ≈ 1.3 × 10⁻⁷ s

2. Example 2Calculator allowed

   Direction of the force on an electron (classic trap)

   An electron moves in the +x direction through a magnetic field that points in the +y direction. Which way is the magnetic force on it?

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   1. 1Step 1: Find $\vec{v} \times \vec{B}$: $\hat{i} \times \hat{j} = \hat{k}$, so it points in +z. (Right hand: fingers along +x, curl toward +y, thumb points +z.)
   2. 2Step 2: $\vec{F} = q\vec{v} \times \vec{B}$ with q negative, so the force points opposite to $\vec{v} \times \vec{B}$.
   3. 3Step 3: The trap is stopping at the right-hand rule and forgetting the electron's negative charge.

   Answer: In the −z direction

3. Example 3Calculator allowed

   Hall voltage in a copper strip

   A copper strip 2.0 cm wide and 1.0 mm thick carries a 10 A current along its length. A 1.5 T magnetic field points perpendicular to its flat face. Copper has 8.5 × 10²⁸ free electrons per cubic meter. Find the Hall voltage across the strip's width.

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   1. 1Step 1: Drift velocity from 11.1, with cross-sectional area (0.020 m)(0.0010 m) = 2.0 × 10⁻⁵ m²: $v_d = \dfrac{I}{nqA} = \dfrac{10}{(8.5 \times 10^{28})(1.60 \times 10^{-19})(2.0 \times 10^{-5})} \approx 3.7 \times 10^{-5}\text{ m/s}$.
   2. 2Step 2: At balance, $qE = qv_dB$, and the field across the width gives $\Delta V_H = Ew = v_dBw$.
   3. 3Step 3: $\Delta V_H = (3.7 \times 10^{-5})(1.5)(0.020) \approx 1.1 \times 10^{-6}\text{ V}$.

   Answer: About 1.1 × 10⁻⁶ V (1.1 μV)

Common mistakes

• !Using the right-hand rule for an electron and forgetting to reverse the answer.
• !Saying a magnetic field speeds up or slows down a charge. The force is perpendicular to the velocity, so it does no work and the speed stays the same.
• !Using the angle between the force and the field in sin θ. The angle is between the velocity and the field.
• !Assuming a faster particle circles more quickly. Its radius grows, but its period doesn't change.

On the exam

• ✓Expect questions that show a particle entering a field region and ask which path it follows, or how the radius changes if the mass, charge, speed or field changes.
• ✓Free-response questions often combine this with energy: a charge is accelerated through a potential difference (9.3), so $q\Delta V = \tfrac{1}{2}mv^2$, then enters a magnetic field. Find v first, then r.
• ✓When you justify a direction, name the rule you used and account for the sign of the charge.