The Workbook

\(\vec F = I\vec L \times \vec B\), where \(\vec L\) points along the current. Magnitude \(BIL\sin\theta\). Perpendicular: \(F = 0.2 \times 3 \times 0.5 = 0.3\) N.

1. \(\vec L\) to the right (\(\hat i\)), \(\vec B\) into the page (\(-\hat k\)): \(\hat i \times (-\hat k) = \hat j\). The force is up the page.

2. \(F = BIL = 0.2 \times 3 \times 0.5 = 0.3\) N.

\(F = BIL\sin 30^\circ = 0.3 \times 0.5 = 0.15\) N. Parallel to the field, \(\sin\theta = 0\): no force.

Each electron feels \(ev_dB\). The length \(L\) contains \(nAL\) electrons, so the total force is \(nALev_dB\). The current is \(I = nAev_d\), so \(F = (nAev_d)LB = BIL\). The forces on the electrons are passed to the wire through collisions with the lattice.

\(\vec F = I\vec L \times \vec B = 2 \times 0.5 \times 0.4\,(\hat j \times \hat i) = 0.4(-\hat k) = -0.4\hat k\) N.

The magnetic force must balance the weight: \(BIL = mg\), so \(I = \dfrac{mg}{BL} = \dfrac{0.1 \times 10}{0.5 \times 1} = 2\) A. The force must be upward, which fixes the direction of the current by \(\vec L \times \vec B\).

In a uniform field, a wire of any shape feels the same force as a straight wire joining its two ends. So \(\vec F = I\vec L_{\text{eff}} \times \vec B\). For a closed loop the ends coincide, \(\vec L_{\text{eff}} = 0\), and the net force is zero.

1. The effective length is the diameter \(AB\): \(2R = 0.2\) m.

2. \(F = BIL_{\text{eff}} = 0.2 \times 5 \times 0.2 = 0.2\) N, perpendicular to \(AB\) in the plane of the page.

1. The ends are \(0.3 + 0.2 + 0.3 = 0.8\) m apart along a straight line, so \(L_{\text{eff}} = 0.8\) m.

2. \(F = 0.2 \times 5 \times 0.8 = 0.8\) N, perpendicular to the line joining the ends.

The loop is closed, so the net force is zero. The two sides perpendicular to the field feel equal and opposite forces along different lines, forming a couple. So there is a torque that tends to turn the loop's plane perpendicular to the field.

\(\vec L_{\text{eff}}\) is the diameter \(AB\), \(0.2\) m. With \(\vec B\) in the plane and perpendicular to \(AB\): \(F = BIL_{\text{eff}} = 0.2 \times 5 \times 0.2 = 0.2\) N, directed perpendicular to the plane of the semicircle. With \(\vec B\) along \(AB\), \(\vec L_{\text{eff}} \parallel \vec B\) and the force is zero.

Split the loop into small elements \(d\vec l\). The total force is \(I\left(\sum d\vec l\right) \times \vec B\). For a closed loop the vector sum of the elements is zero, because the path returns to its start. So the net force is zero whatever the shape.

\(\dfrac{F}{L} = \dfrac{\mu_0I_1I_2}{2\pi d} = \dfrac{2 \times 10^{-7} \times 10 \times 20}{0.1} = 4 \times 10^{-4}\) N/m.

1. At the upper wire the field of the lower wire (current to the right) points out of the page. The force \(I\vec L \times \vec B\) on the upper current, to the right, is then downward, towards the lower wire. Likewise the lower wire is pulled up. Same direction: attraction.

2. Yes. Newton's third law holds: \(\dfrac{\mu_0I_1I_2}{2\pi d}\) per unit length on each, in opposite directions.

Let the same steady current flow in two infinitely long parallel wires one metre apart in vacuum. The current is one ampere when the force between them is \(2 \times 10^{-7}\) newton per metre of length.

Middle wire: pulled equally towards each neighbour, net zero. Outer wire: attracted by the middle wire, \(\dfrac{2 \times 10^{-7} \times 100}{0.1} = 2 \times 10^{-4}\) N/m, and by the far wire, \(\dfrac{2 \times 10^{-7} \times 100}{0.2} = 10^{-4}\) N/m, both inward: \(3 \times 10^{-4}\) N/m towards the centre.

1. Near side (\(0.05\) m, currents parallel): \(F = \dfrac{2 \times 10^{-7} \times 20 \times 10 \times 0.2}{0.05} = 1.6 \times 10^{-4}\) N, towards the wire. Far side (\(0.1\) m, currents antiparallel): \(8 \times 10^{-5}\) N, away from the wire.

2. The short sides carry opposite currents and sit in the same field pattern, so their forces are equal and opposite and cancel.

3. Net \(= 1.6 \times 10^{-4} - 8 \times 10^{-5} = 8 \times 10^{-5}\) N, towards the wire: the loop is attracted.

\(\dfrac{F}{L} = \dfrac{2 \times 10^{-7} \times 100 \times 100}{0.1} = 0.02\) N/m. Opposite currents repel.

\(F = BIL = 0.5 \times 4 \times 0.2 = 0.4\) N. \(W = Fs = 0.4 \times 0.1 = 0.04\) J. The energy comes from the source driving the current.

The horizontal magnetic force must reach the limiting friction: \(BIL = \mu mg = 0.2 \times 2 = 0.4\) N. \(I = \dfrac{0.4}{0.5 \times 0.5} = 1.6\) A.

Parallel: \(\sin 0^\circ = 0\), no force. Perpendicular: the maximum \(BIL\). The perpendicular component of the wire to the field is what counts.

\(F = BIL\sin\theta = 0.1 \times 10 \times 1 \times 0.6 = 0.6\) N.

\(\dfrac{F}{L} = \dfrac{2 \times 10^{-7} \times 1 \times 1}{1} = 2 \times 10^{-7}\) N/m. This is exactly the force used to define the ampere.

Take the diameter along \(x\) and the arc above it. An element at angle \(\theta\) from the \(x\) axis has length \(R\,d\theta\) and feels \(dF = BIR\,d\theta\) directed radially outward. The \(x\) components cancel in pairs; the \(y\) components add: \(F_y = \int_0^\pi BIR\sin\theta\,d\theta = 2BIR\). It acts along the perpendicular bisector of the diameter, the same as for a straight wire of length \(2R\).

\(M = NIA\), directed along the normal to the loop in the sense given by curling the fingers of the right hand along the current. \(M = 100 \times 0.5 \times 0.02 = 1\) A m\(^2\).

1. Each \(0.2\) m side is perpendicular to \(\vec B\): \(F = BIL = 0.5 \times 2 \times 0.2 = 0.2\) N. The currents in the two sides are opposite, so the forces are opposite, one out of the page and one into it. They act along different lines: a couple.

2. Torque \(=\) force \(\times\) separation \(= 0.2 \times 0.1 = 0.02\) N m. Check: \(NIAB = 1 \times 2 \times 0.02 \times 0.5 = 0.02\) N m.

\(\vec M\) is now along \(\vec B\) (or opposite to it), so \(\tau = MB\sin\theta = 0\). With \(\vec M \parallel \vec B\) the energy is least and a small turn brings a restoring torque: stable. With \(\vec M\) antiparallel the energy is greatest and any turn grows: unstable.

\(\vec\tau = \vec M \times \vec B = 2 \times 0.3\,(\hat i \times \hat j) = 0.6\hat k\) N m.

Let the sides of length \(b\) be perpendicular to \(\vec B\). Each feels \(F = BIb\), in opposite directions, forming a couple. The perpendicular distance between their lines of action is \(a\sin\theta\), where \(\theta\) is the angle between the normal and \(\vec B\). Torque \(= BIb \times a\sin\theta = BIA\sin\theta\). The other two sides feel forces along the axis that cancel.

\(M = NIA = 50 \times 1 \times \pi(0.1)^2 = 0.5\pi\) A m\(^2\). \(\tau = MB\sin 30^\circ = 0.5\pi \times 0.2 \times 0.5 = 0.05\pi \approx 0.16\) N m.

Opposite sides carry opposite currents in the same field, so their forces are equal and opposite and the net force is zero. But the two forces act along different lines, so they form a couple, and a couple has a torque without a net force.

\(U = -\vec M \cdot \vec B = -MB\cos\theta\). Parallel: \(U = -0.5\) J. Antiparallel: \(U = +0.5\) J.

(i) \(W = U(90^\circ) - U(0) = 0 - (-MB) = MB = 0.5\) J. (ii) \(W = MB - (-MB) = 2MB = 1\) J.

1. \(P\) (\(\theta = 0\), \(U = -MB\)) is a minimum of energy. A small displacement raises \(U\) and the torque brings the dipole back, so it is stable. \(Q\) (\(\theta = 180^\circ\), \(U = +MB\)) is a maximum: unstable.

2. \(\tau = MB\sin\theta\) is largest at \(\theta = 90^\circ\), where the slope of the \(U\) curve is steepest.

At \(90^\circ\) the torque is greatest and turns the coil towards \(\theta = 0\). It overshoots, is turned back, and oscillates about the parallel position, like a compass needle. With no damping the oscillation continues; in practice it dies down and the coil settles with \(\vec M \parallel \vec B\).

\(T = 2\pi\sqrt{\dfrac{2 \times 10^{-4}}{1 \times 0.5}} = 2\pi\sqrt{4 \times 10^{-4}} = 2\pi \times 0.02 = 0.04\pi \approx 0.13\) s.

1. Cylindrical pole faces, together with the iron core, make the field lines in the gap point along radii: a radial field.

2. Whichever way the coil turns, its plane stays parallel to the local field and its normal stays perpendicular to \(\vec B\). The angle in \(\tau = NIAB\sin\theta\) is always \(90^\circ\).

Being soft iron, it concentrates the field lines through the gap, giving a strong field and so a large torque. Its curved surface, matching the pole pieces, makes the field radial so that the deflection is proportional to the current.

Deflecting torque \(= \) restoring torque: \(NIAB = k\theta\), so \(\theta = \dfrac{NAB}{k}I = \dfrac{100 \times 2 \times 10^{-4} \times 0.5}{10^{-4}}I = 100I\) rad per ampere. For \(1\) mA: \(0.1\) rad, about \(6^\circ\).

Current sensitivity \(\dfrac{\theta}{I} = \dfrac{NAB}{k}\). Voltage sensitivity \(\dfrac{\theta}{V} = \dfrac{NAB}{kR}\). Doubling \(N\) doubles the wire length and so doubles \(R\); the ratio \(\dfrac{N}{R}\) is unchanged and the voltage sensitivity stays the same.

1. \(I = \dfrac{k\theta}{NAB} = \dfrac{10^{-4} \times 0.2}{10^{-2}} = 2 \times 10^{-3}\) A \(= 2\) mA.

2. \(\theta = 100 \times 0.5 \times 10^{-3} = 0.05\) rad.

Phosphor bronze has a small torsion constant, so a small torque gives a large deflection. It is elastic and does not take a permanent twist. It is non-magnetic, and it conducts, so it can carry the current into the coil.

The figure of merit is the current needed for a deflection of one division. Here \(\dfrac{3 \text{ mA}}{30} = 0.1\) mA per division.

In the radial field \(\theta = \dfrac{NAB}{k}I\), so equal steps of current give equal steps of deflection. In a uniform field the torque would be \(NIAB\cos\theta\) (with \(\theta\) the deflection), giving \(k\theta = NIAB\cos\theta\): a non-linear, cramped scale at large deflections.

1. \(A = 0.04 \times 0.05 = 2 \times 10^{-3}\) m\(^2\). \(M = NIA = 40 \times 3 \times 2 \times 10^{-3} = 0.24\) A m\(^2\).

2. \(\tau = MB\sin 37^\circ = 0.24 \times 0.25 \times 0.6 = 0.036\) N m.

\(\tau_{\max} = MB = 2 \times 0.1 = 0.2\) N m, at \(90^\circ\). Energy difference \(= MB - (-MB) = 2MB = 0.4\) J.

The orbiting electron is a current \(I = \dfrac{ev}{2\pi r}\) round an area \(\pi r^2\), so \(M = \dfrac{evr}{2} = \dfrac{1.6 \times 10^{-19} \times 2.2 \times 10^{6} \times 5.3 \times 10^{-11}}{2} \approx 9.3 \times 10^{-24}\) A m\(^2\), the Bohr magneton.

In a radial field the plane of the coil is always parallel to \(\vec B\). The deflecting torque is then \(NIAB\) for every position, with no \(\sin\theta\) factor. The spring supplies \(k\theta\). At rest \(NIAB = k\theta\), so \(\theta = \dfrac{NAB}{k}I \propto I\).