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Force on a moving charge questions
The same rule applies to a single moving charge as to a current-carrying wire: F = BQv sin θ, at right angles to both the velocity and the field. A force always perpendicular to the motion does no work, so it changes direction without changing speed, which sends the charge in a circle. That is the principle of the cyclotron.
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 kgExplain why the magnetic force on a moving charged particle does no work on the particle.
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.
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.
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.
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 kgShow 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.
Describe how a cyclotron uses magnetic and electric fields to accelerate charged particles.
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 kgIn 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 kgA 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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