10 exam-style questions on force on currents and charges, magnetic flux, Faraday’s and Lenz’s laws. Tap Show answer for full working.
A straight wire of length \(0.25\,\mathrm{m}\) carries a current of \(3.0\,\mathrm{A}\) perpendicular to a magnetic field of flux density \(0.40\,\mathrm{T}\). Calculate the force on the wire.
\(F = BIl = 0.40\times3.0\times0.25 = 0.30\,\mathrm{N}\).
An electron moves at \(2.0\times10^{6}\,\mathrm{m\,s}^{-1}\) perpendicular to a magnetic field of \(0.15\,\mathrm{T}\). Calculate the magnitude of the force on the electron. (\(e = 1.6\times10^{-19}\,\mathrm{C}\).)
\(F = Bev = 0.15\times1.6\times10^{-19}\times2.0\times10^{6} = 4.8\times10^{-14}\,\mathrm{N}\).
State Faraday’s law and Lenz’s law.
Faraday: the magnitude of the induced EMF is equal to the rate of change of magnetic flux linkage (\(\mathcal{E} = -N\mathrm{d}\Phi/\mathrm{d}t\)).
Lenz: the direction of the induced EMF is such that it opposes the change that produces it.
A circular coil of \(150\) turns and radius \(4.0\,\mathrm{cm}\) is placed with its plane perpendicular to a magnetic field that decreases from \(0.60\,\mathrm{T}\) to \(0.20\,\mathrm{T}\) in \(0.050\,\mathrm{s}\). Calculate the average induced EMF.
\(\Phi = BA = 0.60\times\pi\times(0.04)^{2}\) initially; \(\Delta\Phi = 0.40\times\pi\times0.0016 = 2.01\times10^{-3}\,\mathrm{Wb}\).
\(\mathcal{E} = N\Delta\Phi/\Delta t = 150\times2.01\times10^{-3}/0.050 = 6.03\,\mathrm{V}\).
A proton moves in a circular path of radius \(0.50\,\mathrm{m}\) in a magnetic field of \(0.80\,\mathrm{T}\). Calculate the speed of the proton. (Proton mass \(= 1.67\times10^{-27}\,\mathrm{kg}\), \(e = 1.6\times10^{-19}\,\mathrm{C}\).)
\(r = mv/(Be) \Rightarrow v = Ber/m = 0.80\times1.6\times10^{-19}\times0.50/1.67\times10^{-27} = 3.83\times10^{7}\,\mathrm{m\,s}^{-1}\).
A rectangular coil of \(80\) turns, area \(0.020\,\mathrm{m}^{2}\), rotates at \(50\,\mathrm{Hz}\) in a magnetic field of \(0.15\,\mathrm{T}\). Calculate the maximum induced EMF.
\(\mathcal{E}_{\max} = NBA\omega = 80\times0.15\times0.020\times2\pi\times50 = 75.4\,\mathrm{V}\).
Explain why a conducting rod moving perpendicular to a magnetic field has an induced EMF between its ends, and state the factors that determine the magnitude of the EMF.
Free charges in the rod experience a magnetic force \(F = Bqv\) that drives them to one end, creating a potential difference until the electric force balances the magnetic force.
Magnitude \(\mathcal{E} = Blv\) depends on field strength, length of rod and speed of motion (all perpendicular components).
A coil of \(200\) turns and area \(0.015\,\mathrm{m}^{2}\) has a resistance of \(4.0\,\Omega\). It is placed in a magnetic field of \(0.50\,\mathrm{T}\) that is reduced to zero in \(0.10\,\mathrm{s}\). Calculate the average induced current and the charge that flows through the coil.
\(\mathcal{E}_{\mathrm{avg}} = N\Delta\Phi/\Delta t = 200\times0.50\times0.015/0.10 = 15\,\mathrm{V}\).
\(I_{\mathrm{avg}} = \mathcal{E}/R = 15/4 = 3.75\,\mathrm{A}\).
\(Q = I\Delta t = 3.75\times0.10 = 0.375\,\mathrm{C}\) (or \(Q = N\Delta\Phi/R\)).
A long straight wire carries a current of \(8.0\,\mathrm{A}\). Calculate the magnetic flux density at a distance of \(4.0\,\mathrm{cm}\) from the wire. A second parallel wire \(4.0\,\mathrm{cm}\) away carries \(5.0\,\mathrm{A}\) in the same direction. Calculate the force per unit length on each wire and state whether it is attractive or repulsive. (\(\mu_{0} = 4\pi\times10^{-7}\).)
\(B = \mu_{0}I/(2\pi r) = 4\pi\times10^{-7}\times8/(2\pi\times0.04) = 4.0\times10^{-5}\,\mathrm{T}\).
\(F/l = \mu_{0}I_{1}I_{2}/(2\pi d) = 4\pi\times10^{-7}\times8\times5/(2\pi\times0.04) = 2.0\times10^{-4}\,\mathrm{N\,m}^{-1}\) (attractive — currents in same direction).
A square loop of side \(5.0\,\mathrm{cm}\) and resistance \(0.20\,\Omega\) moves at constant speed \(2.0\,\mathrm{m\,s}^{-1}\) into a uniform magnetic field of \(0.30\,\mathrm{T}\) directed into the page. The field region is wide. Calculate the induced current while the loop is entering the field, and sketch a graph of induced current against time from the moment the leading edge enters until the trailing edge leaves a field region \(20\,\mathrm{cm}\) wide.
While entering: \(\mathcal{E} = B\ell v = 0.30\times0.05\times2.0 = 0.030\,\mathrm{V}\); \(I = 0.030/0.20 = 0.15\,\mathrm{A}\).
Current is constant while entering (0.025 s), zero while fully inside, then constant (opposite direction) while leaving. Graph: rectangular pulse positive, gap, rectangular pulse negative.