D.3 Motion in electromagnetic fields

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Force on a moving charge

F = qvB\sin\theta

Force on a conductor

F = BIL\sin\theta

Radius in a magnetic field

r = \frac{mv}{qB}

Parallel wires (per length)

\frac{F}{L} = \mu_{0}\frac{I_{1}I_{2}}{2\pi r}

Velocity selector

v = \frac{E}{B}

Section A — Multiple choice

A1. A magnetic force never changes the kinetic energy of a charged particle because:

A The force is too small
B The force is always perpendicular to the velocity, so it does no work
C Magnetic fields do not exert forces on charges
D Kinetic energy is conserved in all fields

A2. An electron and a proton enter the same magnetic field with the same velocity. Compared with the proton, the electron's circular path has:

A A much smaller radius, curving the opposite way
B A much larger radius, curving the opposite way
C The same radius, curving the same way
D A much smaller radius, curving the same way

A3. Two long parallel wires carry currents in the same direction. They:

A Repel each other
B Attract each other
C Exert no force on each other
D Rotate to become perpendicular

Section B — Short answer

B1. An electron ( $m = 9.11\times10^{-31}$ kg) moving at $2.0\times10^7\ \text{m s}^{-1}$ enters a 0.50 mT field perpendicularly. Calculate the radius of its path. [3 marks]

B2. In a velocity selector, an electric field of $3.0\times10^4\ \text{V m}^{-1}$ is crossed with a magnetic field of 0.060 T. Determine the speed of ions that pass undeflected, and state what happens to faster ions. [3 marks]

B3. A horizontal wire of length 0.30 m carrying 4.0 A lies perpendicular to a horizontal 0.25 T field. Calculate the force on the wire, and state the condition for the wire to "float". [3 marks]

Answer key

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Section A

A1: The force is always perpendicular to the velocity, so it does no work — $W = Fs\cos\theta$ with $\theta = 90°$ gives zero at every instant: the force steers without speeding or slowing. Electric fields, by contrast, do work and change kinetic energy.

A2: A much smaller radius, curving the opposite way — $r = mv/qB$ : same speed and charge magnitude, but the electron's mass is ~1836 times smaller, so its radius shrinks by the same factor. Opposite charge sign reverses the force direction, hence the opposite sense of rotation.

A3: Attract each other — Each wire sits in the magnetic field of the other; applying $F = BIL$ with the field directions shows attraction for parallel currents and repulsion for antiparallel. This force defined the ampere.

Section B

B1: $r = \frac{mv}{qB} = \frac{(9.11\times10^{-31})(2.0\times10^7)}{(1.6\times10^{-19})(5.0\times10^{-4})} \approx 0.23\ \text{m}$ . Weak fields and fast electrons give conveniently laboratory-sized circles.

B2: Undeflected means $qE = qvB$ , so $v = E/B = 3.0\times10^4/0.060 = 5.0\times10^5\ \text{m s}^{-1}$ . Faster ions experience a magnetic force ( $qvB$ ) exceeding the electric force and deflect towards the magnetic-force side.

B3: $F = BIL = 0.25 \times 4.0 \times 0.30 = 0.30\ \text{N}$ . It floats if this force is directed vertically upward (right-hand/motor rule) and equals the weight: $BIL = mg$ , i.e. the wire's mass must be $0.30/9.81 \approx 31$ g.