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Questions on force on a moving charge
A moving charge in a magnetic field feels a force F = BQv at right angles to its velocity, so it does no work and the speed never changes. The path curves into a circle of radius r = mv/BQ. The Hall probe and the velocity selector balance that force against an electric one, and the cyclotron steers with it.
19 original questions · 52 marks · the force on a moving charge notes · Magnetic fields
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State the equation for the magnetic force on a charge moving through a magnetic field and the condition under which it applies.
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F = BQv, where Q is the charge and v its speed (1); valid when the velocity is perpendicular to the field (1).A particle of charge 1.6 × 10−19 C moves at 2.0 × 106 m s−1 at right angles to a magnetic field of flux density 0.40 T. Calculate the force on the particle.
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F = BQv = 0.40 × 1.6 × 10−19 × 2.0 × 106 (1)
F = 1.28 × 10−13 N (1)Explain why a charged particle moving at right angles to a uniform magnetic field moves in a circle.
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The magnetic force is always perpendicular to the velocity, so it changes the direction of motion but not the speed (1); a force of constant magnitude always at right angles to the velocity acts as a centripetal force, producing circular motion (1).State two situations in which a charged particle in a magnetic field experiences no magnetic force.
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The particle is stationary, since F = BQv is zero when v = 0 (1); the particle moves parallel (or antiparallel) to the field, so it has no component of velocity at right angles to the field (1).An electron travels horizontally due north through a region where the magnetic field is directed vertically downwards. Determine the direction of the magnetic force on the electron.
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The electron is negative, so the conventional current is due south, opposite to its velocity (1); Fleming's left hand rule (field down, current south) gives a force that is horizontal, due east (1).The speed of a charged particle moving at right angles to a uniform magnetic field is doubled. State and explain the effect on the magnetic force on the particle and on the radius of its path.
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F = BQv, so the force doubles (1); r = mv/BQ, so the radius also doubles (1).An electron moves at 3.0 × 106 m s−1 perpendicular to a magnetic field of flux density 0.50 T. Calculate the radius of its circular path.
me = 9.11 × 10−31 kg, e = 1.60 × 10−19 CMark scheme
BQv = mv2/r, so r = mv/BQ (1)
r = (9.11 × 10−31 × 3.0 × 106)/(0.50 × 1.60 × 10−19) (1)
r = 3.42 × 10−5 m (1)A proton moves at 5.0 × 105 m s−1 perpendicular to a magnetic field of flux density 0.80 T. Calculate the magnetic force on the proton.
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F = BQv = 0.80 × 1.60 × 10−19 × 5.0 × 105 (1)
F = 6.40 × 10−14 N (1)Calculate the radius of the circular path of the proton in the previous question.
mp = 1.67 × 10−27 kgMark scheme
r = mv/BQ = (1.67 × 10−27 × 5.0 × 105)/(0.80 × 1.60 × 10−19) (1)
r = 6.52 × 10−3 m (1)Explain why the magnetic force on a moving charged particle does no work on the particle.
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The force is always perpendicular to the velocity (direction of motion) (1); work done requires a component of force along the displacement, and here there is none (1); so the speed and kinetic energy of the particle remain constant (1).An alpha particle of charge +3.2 × 10−19 C and mass 6.64 × 10−27 kg travels at 1.5 × 107 m s−1 at right angles to a magnetic field of flux density 250 mT. Calculate the radius of its circular path.
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r = mv/BQ (1)
= (6.64 × 10−27 × 1.5 × 107)/(0.25 × 3.2 × 10−19) (1)
r = 1.25 m (1)In a velocity selector, an electric field of strength 4.5 × 104 V m−1 is perpendicular to a magnetic field of flux density 0.30 T. State the condition for a charged particle to pass through undeflected and calculate the speed at which this occurs.
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The electric force EQ balances the magnetic force BQv (1)
v = E/B = (4.5 × 104)/0.30 (1)
v = 1.5 × 105 m s−1 (1)In a particle detector, a particle of charge 1.60 × 10−19 C leaves a circular track of radius 45 mm in a magnetic field of flux density 1.2 T. Calculate the momentum of the particle.
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BQv = mv2/r gives momentum p = mv = BQr = 1.2 × 1.60 × 10−19 × 0.045 (1)
p = 8.64 × 10−21 kg m s−1 (1)An electron is accelerated from rest through a potential difference of 500 V and then enters a magnetic field of flux density 0.30 T at right angles to the field. Calculate the speed of the electron and the radius of its circular path.
e = 1.60 × 10−19 C, me = 9.11 × 10−31 kgMark scheme
Ek = eV = 8.00 × 10−17 J (1)
v = √(2Ek/m) = 1.33 × 107 m s−1 (1)
r = mv/BQ (1)
r = 2.52 × 10−4 m (1)Show that the period of the circular motion of a charged particle in a magnetic field is T = 2πm/BQ, for speeds well below the speed of light. Calculate the period and the frequency of this motion for an electron in a field of flux density 0.50 T.
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v = 2πr/T with r = mv/BQ gives T = 2πm/BQ, independent of speed while m may be taken as constant, that is at non-relativistic speeds (1)
T = (2π × 9.11 × 10−31)/(0.50 × 1.60 × 10−19) (1)
T = 7.15 × 10−11 s (1)
f = 1/T = 1.40 × 1010 Hz (1)Describe how a cyclotron uses magnetic and electric fields to accelerate charged particles.
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A uniform magnetic field, perpendicular to the plane of motion, bends the particles into semicircular paths inside two hollow dees (1). An alternating pd across the gap accelerates the particles each time they cross between the dees (1). As the speed increases the radius increases, so the particles spiral outwards until they exit at high energy (1).A particle of charge +1.60 × 10−19 C enters a detector at 3.0 × 106 m s−1, at right angles to a magnetic field of flux density 0.85 T, and follows a circular arc of radius 24 mm. Deduce whether the particle could be a proton.
mp = 1.67 × 10−27 kgMark scheme
r = mv/BQ rearranged: m = BQr/v (1)
m = (0.85 × 1.60 × 10−19 × 0.024)/(3.0 × 106) (1)
m = 1.09 × 10−27 kg (1)
This is only about two thirds of the proton mass of 1.67 × 10−27 kg, so the particle is not a proton (1)In a cyclotron, protons reach their maximum speed at the outer edge of the dees, where the radius of their path is 0.60 m and the flux density is 1.5 T. Calculate the maximum speed of the protons and their maximum kinetic energy.
e = 1.60 × 10−19 C, mp = 1.67 × 10−27 kgMark scheme
r = mv/BQ rearranged: v = BQr/m = (1.5 × 1.60 × 10−19 × 0.60)/(1.67 × 10−27) (1)
v = 8.62 × 107 m s−1 (1)
Ek = ½mv2 (1)
Ek = 6.21 × 10−12 J (1)A Hall probe is used to measure magnetic flux density. Explain how a steady pd, proportional to the flux density, appears across a current-carrying slice of semiconductor placed in the field.
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The charge carriers drifting through the slice move through the field, so each feels a magnetic force BQv that pushes it towards one face (1). Charge builds up on that face until the electric force it produces balances the magnetic force, leaving a steady pd, the Hall voltage, across the faces (1). The Hall voltage is proportional to B, so a calibrated probe reads the flux density directly (1).
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