Moving Charges and Magnetism — Questions & Answers

Sample questions from Moving Charges and Magnetism

1. Q1. A circular coil of radius R carries current I and produces field B0 at its centre. The field at a point on its axis at a distance R from the centre is:

   • A.B0/(2√2)✓
   • B.B0/2
   • C.B0/√2
   • D.B0/4
   Solution

   B(x) = μ0 I R^2/[2(R^2 + x^2)^(3/2)]. At x = R: B = μ0 I R^2/[2 (2R^2)^(3/2)] = μ0 I R^2/(2 × 2√2 R^3) = [μ0 I/(2R)] × 1/(2√2) = B0/(2√2) ≈ 0.354 B0. The field falls off faster than the naive 1/2 because of the 3/2 power.

2. Q2. The magnetic force on a charged particle moving PARALLEL to the magnetic field direction is:

   • A.Maximum possible
   • B.Zero✓
   • C.Equal to qvB
   • D.Equal to qv/B
   Solution

   F = qvB sin(theta), where theta is the angle between velocity and field. When the particle moves parallel to B, theta=0, so sin(theta)=0 and the force is zero. A student who forgets the angle dependence and always uses F=qvB picks C.

3. Q3. A tightly wound toroid (all turns closely spaced, negligible pitch) is essentially:

   • A.Only usable for measuring electric fields, not magnetic fields
   • B.Identical in its external field to a bar magnet
   • C.A device with zero magnetic field anywhere, including inside
   • D.A solenoid bent in the shape of a circle, confining the field entirely within the toroid's core✓
   Solution

   A toroid can be thought of as a straight solenoid bent around in a closed circular shape, so that the field lines (which would otherwise fringe out at a solenoid's ends) are confined entirely within the toroid's core, with essentially zero field outside (in the ideal, tightly-wound case). A student who thinks a toroid behaves externally like a bar magnet (as a finite straight solenoid does) picks B, missing the key difference that the toroid has NO external field due to its closed-loop geometry.

4. Q4. A cyclotron is designed for protons with a magnetic field of 1T. If the SAME cyclotron (same B) is instead used to accelerate deuterons (charge +e, same as proton, but mass about twice the proton's mass), the required oscillator frequency, compared to the proton case, must be:

   • A.Halved (since f=qB/2 pi m and m has doubled, q unchanged)✓
   • B.Doubled
   • C.Unchanged, since B is unchanged
   • D.Quadrupled
   Solution

   Since f = qB/(2 pi m), and the deuteron has the SAME charge q as the proton but roughly DOUBLE the mass, the required frequency is halved (f is inversely proportional to m, with q and B fixed). A student who thinks frequency depends only on B (forgetting the mass and charge dependence) picks C.

5. Q5. A charged particle enters a region of crossed fields with E = 1.0 x 10^4 V/m and B = 0.20 T mutually perpendicular. A particle passing undeviated must have speed:

   • A.5.0 x 10^4 m/s✓
   • B.2.0 x 10^3 m/s
   • C.5.0 x 10^3 m/s
   • D.2.0 x 10^5 m/s
   Solution

   Balancing qE against qvB gives v = E/B = 1.0 x 10^4/0.20 = 5.0 x 10^4 m/s, independent of the charge and mass of the particle. Multiplying E by B instead of dividing gives 2.0 x 10^3, which has the wrong dependence entirely.

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