Magnetism and Magnetic Effects: Common Mistakes and Fixes (9)

hard • 3 min read
Tags Magnetism

Question

A proton enters a uniform magnetic field B=0.5B = 0.5 T moving with velocity v=2×106v = 2 \times 10^6 m/s perpendicular to the field. Find the radius of its circular path and the time period of revolution. Mass of proton mp=1.67×10−27m_p = 1.67 \times 10^{-27} kg, charge q=1.6×10−19q = 1.6 \times 10^{-19} C.

Solution — Step by Step

When a charged particle enters a uniform B⃗\vec{B} field perpendicular to its velocity, the magnetic force F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B} is always perpendicular to v⃗\vec{v}. This causes uniform circular motion.

qvB=mv2rqvB = \frac{mv^2}{r}

r=mvqBr = \frac{mv}{qB}

r=(1.67×10−27)(2×106)(1.6×10−19)(0.5)r = \frac{(1.67 \times 10^{-27})(2 \times 10^6)}{(1.6 \times 10^{-19})(0.5)}

r=3.34×10−210.8×10−19=4.175×10−2 m≈4.18 cmr = \frac{3.34 \times 10^{-21}}{0.8 \times 10^{-19}} = 4.175 \times 10^{-2} \text{ m} \approx 4.18 \text{ cm}

T=2πrv=2πmqBT = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

T=2π(1.67×10−27)(1.6×10−19)(0.5)=1.05×10−260.8×10−19≈1.31×10−7 sT = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.5)} = \frac{1.05 \times 10^{-26}}{0.8 \times 10^{-19}} \approx 1.31 \times 10^{-7} \text{ s}

Final answer: r≈4.18r \approx 4.18 cm, T≈1.31×10−7T \approx 1.31 \times 10^{-7} s.

Why This Works

The magnetic force does no work on a moving charge (because F⃗⊥v⃗\vec{F} \perp \vec{v}), so the speed stays constant. With constant speed and a force always perpendicular to velocity, the path is a circle.

The crucial observation: T=2πm/(qB)T = 2\pi m/(qB) does not depend on vv or rr. This is the principle behind the cyclotron — particles of all energies (in the non-relativistic regime) take the same time per loop.

Radius: r=mvqB=pqBr = \dfrac{mv}{qB} = \dfrac{p}{qB}

Period: T=2πmqBT = \dfrac{2\pi m}{qB} (independent of speed)

Frequency: f=qB2πmf = \dfrac{qB}{2\pi m} (cyclotron frequency)

Alternative Method

Once you know TT is independent of vv, compute TT first, then r=vT/(2π)r = vT/(2\pi):

r=(2×106)(1.31×10−7)2π≈4.18 cm✓r = \frac{(2 \times 10^6)(1.31 \times 10^{-7})}{2\pi} \approx 4.18 \text{ cm} \checkmark

Nine classic traps in magnetism:

  1. Using F=qEF = qE instead of F=qvBF = qvB. EE is electric field, BB is magnetic.
  2. Forgetting the sin θ when v⃗\vec{v} is not perpendicular to B⃗\vec{B}. General form: F=qvBsin⁡θF = qvB\sin\theta.
  3. Mixing up Lorentz force directions — use right-hand rule for positive charges, left for electrons.
  4. Treating helical motion as planar. If v⃗\vec{v} has a component along B⃗\vec{B}, the particle moves in a helix.
  5. Forgetting that TT is independent of vv — leads to wrong cyclotron design questions.
  6. Using radius formula r=mv/qBr = mv/qB with non-SI units (e.g., gauss instead of tesla).
  7. Confusing BB (magnetic flux density, T) with HH (magnetic field intensity, A/m).
  8. For solenoids: B=μ0nIB = \mu_0 n I where nn is turns per metre, not total turns.
  9. Sign of induced EMF (Lenz’s law) — always opposes the change in flux.

Common Mistake

The biggest one: treating the direction of force as parallel to B⃗\vec{B}. The magnetic force is perpendicular to both v⃗\vec{v} and B⃗\vec{B}. Use F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B} and the right-hand rule — never just F=qBF = qB as a scalar.

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