The Workbook

A current element \(I\,d\vec l\) produces at a point at distance \(r\) a field \(dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}\). Here \(\theta\) is the angle between \(d\vec l\) and the line to the point. The direction is that of \(d\vec l \times \hat r\), perpendicular to both.

\(B = \dfrac{\mu_0I}{2\pi d} = \dfrac{2 \times 10^{-7} \times 10}{0.05} = 4 \times 10^{-5}\) T. Point the right thumb along the current; the curled fingers give the direction of the circular field lines.

1. Each wire gives \(\dfrac{2 \times 10^{-7} \times 10}{0.1} = 2 \times 10^{-5}\) T at \(P\). With parallel currents the two fields at the midpoint are opposite (one into the page, one out), so the net field is zero.

2. With antiparallel currents the two fields at \(P\) point the same way: \(B = 4 \times 10^{-5}\) T.

Half-length \(0.3\) m, \(d = 0.4\) m: \(\sin\alpha = \sin\beta = \dfrac{0.3}{0.5} = 0.6\). \(B = \dfrac{10^{-7} \times 10}{0.4}(0.6 + 0.6) = 3 \times 10^{-6}\) T.

A semi-infinite wire gives \(\dfrac{\mu_0I}{4\pi d}\) at the foot of the perpendicular from its end. Here that is \(\dfrac{10^{-7} \times 10}{0.05} = 2 \times 10^{-5}\) T, exactly half the value for the infinite wire in Q2.

For a point on the line of the wire, every element \(d\vec l\) is parallel to the vector \(\hat r\) to the point. So \(\sin\theta = 0\) and each \(dB\) is zero.

\(B = \dfrac{\mu_0NI}{2R} = \dfrac{4\pi \times 10^{-7} \times 1 \times 5}{0.2} = \pi \times 10^{-5} \approx 3.1 \times 10^{-5}\) T.

1. An arc is a fraction \(\dfrac{\theta}{2\pi}\) of a full loop: \(B = \dfrac{\theta}{2\pi}\cdot\dfrac{\mu_0I}{2R} = \dfrac{\mu_0I\theta}{4\pi R}\).

2. \(B = 10^{-7} \times \dfrac{8 \times \pi/2}{0.1} = 4\pi \times 10^{-6} \approx 1.3 \times 10^{-5}\) T.

1. Zero. \(C\) lies on the line of each straight lead, where \(\sin\theta = 0\) for every element.

2. Only the semicircle counts: \(B = \dfrac{\mu_0I}{4R} = \dfrac{4\pi \times 10^{-7} \times 4}{0.4} = 4\pi \times 10^{-6} \approx 1.3 \times 10^{-5}\) T. The current runs clockwise over the arc as drawn, so the field at \(C\) is into the page.

\(B = \dfrac{\mu_0IR^2}{2(R^2 + x^2)^{3/2}}\). At \(x = \sqrt3R\): \((R^2 + 3R^2)^{3/2} = 8R^3\), so \(B = \dfrac{\mu_0I}{16R} = \dfrac{1}{8}\) of the centre value: \(\dfrac{\pi \times 10^{-5}}{8} \approx 3.9 \times 10^{-6}\) T.

Opposite senses give opposite fields: \(B = \dfrac{\mu_0I}{2}\left(\dfrac{1}{0.1} - \dfrac{1}{0.2}\right) = 2\pi \times 10^{-7} \times 5 \times 5 = 5\pi \times 10^{-6} \approx 1.6 \times 10^{-5}\) T, in the sense of the smaller loop.

1. \(B_1 = \dfrac{\mu_0I}{4\pi d}(\sin 45^\circ + \sin 45^\circ) = \dfrac{10^{-7} \times 5}{0.1} \times \sqrt2 = 5\sqrt2 \times 10^{-6} \approx 7.1 \times 10^{-6}\) T.

2. All four sides give fields in the same sense at \(O\): \(B = 4B_1 = 2\sqrt2 \times 10^{-5} \approx 2.8 \times 10^{-5}\) T.

Circle: \(R = \dfrac{L}{2\pi}\), \(B_c = \dfrac{\mu_0I}{2R} = \dfrac{\pi\mu_0I}{L}\). Square: \(a = \dfrac{L}{4}\), \(B_s = \dfrac{2\sqrt2\mu_0I}{\pi a} = \dfrac{8\sqrt2\mu_0I}{\pi L}\). \(\dfrac{B_s}{B_c} = \dfrac{8\sqrt2}{\pi^2} \approx 1.15\): the square gives a slightly larger field.

1. Zero: \(P\) lies on the line of the horizontal arm.

2. \(P\) is at the foot of the perpendicular from the end \(O\) of the vertical arm, at \(d = 0.1\) m: \(B = \dfrac{\mu_0I}{4\pi d} = \dfrac{10^{-7} \times 20}{0.1} = 2 \times 10^{-5}\) T. That is the whole field at \(P\).

The centre is at \(d = \dfrac{\sqrt3}{2}a\) from each side, and each side subtends \(30^\circ\) on either side of the perpendicular. One side: \(\dfrac{\mu_0I}{4\pi d} \times 2\sin 30^\circ = \dfrac{\mu_0I}{4\pi d}\). Six sides: \(B = \dfrac{6\mu_0I}{4\pi d} = \dfrac{\sqrt3\mu_0I}{\pi a} = \dfrac{1.732 \times 4\pi \times 10^{-7} \times 10}{\pi \times 0.1} \approx 6.9 \times 10^{-5}\) T.

Wire: \(\dfrac{2 \times 10^{-7} \times 10}{0.1} = 2 \times 10^{-5}\) T. Loop: \(\dfrac{\mu_0I}{2R} = \dfrac{4\pi \times 10^{-7} \times 2}{0.1} \approx 2.5 \times 10^{-5}\) T. Opposite directions: net \(\approx 0.5 \times 10^{-5} = 5 \times 10^{-6}\) T, in the direction of the loop's own field.

The midpoint is \(x = 0.1\) m \(= R\) from each loop: \(B_1 = \dfrac{\mu_0IR^2}{2(2R^2)^{3/2}} = \dfrac{B_{\text{centre}}}{2\sqrt2} = \dfrac{\pi \times 10^{-5}}{2.83} \approx 1.1 \times 10^{-5}\) T. Both add: \(B \approx 2.2 \times 10^{-5}\) T.

For an anticlockwise current the right-hand rule gives a field out of the page at the centre. On the axis the field keeps the same direction at every distance, so it points out of the page far away as well, only weaker.

A long wire has current at every distance along it, and the contributions add up to a slowly falling \(\dfrac{1}{r}\) field. A loop is a closed current whose near and far sides nearly cancel far away. It acts as a magnetic dipole, whose field falls as \(\dfrac{1}{x^3}\).

\(B = \dfrac{2 \times 10^{-7} \times 100}{1} = 2 \times 10^{-5}\) T, about \(0.4\) of the Earth's field. A compass near the cable would be noticeably deflected.

\(B = \dfrac{\mu_0NI}{2R}\), so \(I = \dfrac{2RB}{\mu_0N} = \dfrac{2 \times 0.05 \times 10^{-3}}{4\pi \times 10^{-7} \times 100} \approx 0.8\) A.

\(B \propto \dfrac{N}{R}\) at fixed \(I\). Doubling both leaves the ratio, and so the field, unchanged.

\(B = \dfrac{\mu_0I\theta}{4\pi R} = 10^{-7} \times \dfrac{6 \times \pi/3}{0.1} = 2\pi \times 10^{-6} \approx 6.3 \times 10^{-6}\) T.

Every element \(d\vec l\) of the loop is perpendicular to the radius to the centre, so \(\sin\theta = 1\) and \(r = R\): \(dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{R^2}\). All the \(dB\) point the same way, along the axis. Adding round the loop, \(\sum dl = 2\pi R\): \(B = \dfrac{\mu_0}{4\pi}\dfrac{I \times 2\pi R}{R^2} = \dfrac{\mu_0I}{2R}\).

The line integral of \(\vec B\) round any closed loop equals \(\mu_0\) times the net current threading the loop: \(\oint\vec B\cdot d\vec l = \mu_0I_{\text{enc}}\). Curl the fingers of the right hand in the sense of travel round the loop. Current along the thumb counts positive, current the other way negative.

By symmetry the field lines are circles centred on the wire. \(B\) is the same all round a circle of radius \(r\) and points along the circle. So \(\oint\vec B\cdot d\vec l = B \times 2\pi r = \mu_0I\), giving \(B = \dfrac{\mu_0I}{2\pi r}\). The cylindrical symmetry makes \(B\) constant on the loop, which is what the law needs.

1. Uniform current: \(I_{\text{enc}} = I\dfrac{\pi r^2}{\pi R^2} = I\dfrac{r^2}{R^2}\). \(B \times 2\pi r = \mu_0I\dfrac{r^2}{R^2}\), so \(B = \dfrac{\mu_0Ir}{2\pi R^2} = \dfrac{2 \times 10^{-7} \times 20 \times 0.01}{(0.02)^2} = 10^{-4}\) T.

2. \(B = \dfrac{\mu_0I}{2\pi R} = \dfrac{2 \times 10^{-7} \times 20}{0.02} = 2 \times 10^{-4}\) T.

3. Outside, all the current is enclosed: \(B = \dfrac{2 \times 10^{-7} \times 20}{0.04} = 10^{-4}\) T.

A circular loop inside the hollow encloses no current, so \(B \times 2\pi r = 0\) and \(B = 0\) inside. A loop outside encloses the whole current \(I\), so \(B = \dfrac{\mu_0I}{2\pi r}\), the same as for a thin wire on the axis.

1. Inside: \(B = \dfrac{\mu_0Ir}{2\pi R^2}\), rising in proportion to \(r\). Outside: \(B = \dfrac{\mu_0I}{2\pi r}\), falling as \(\dfrac{1}{r}\).

2. The two expressions meet at \(r = R\), where \(B = \dfrac{\mu_0I}{2\pi R}\) is the largest value.

\(I(r) = \int_0^r J_0\dfrac{r'}{R}2\pi r'\,dr' = \dfrac{2\pi J_0r^3}{3R}\). The total current is \(I = I(R) = \dfrac{2\pi J_0R^2}{3}\), so \(I(r) = I\dfrac{r^3}{R^3}\). Ampere: \(B\,2\pi r = \mu_0I\dfrac{r^3}{R^3}\), so \(B = \dfrac{\mu_0Ir^2}{2\pi R^3}\). At \(r = \dfrac{R}{2}\): \(B = \dfrac{\mu_0I}{8\pi R}\), half of the uniform-current value \(\dfrac{\mu_0I}{4\pi R}\), because less of the current lies near the axis.

In the gap a loop encloses only the inner \(5\) A: \(B = \dfrac{2 \times 10^{-7} \times 5}{0.01} = 10^{-4}\) T. Outside, the loop encloses \(5 - 5 = 0\): \(B = 0\). That is why coaxial cables do not disturb their surroundings.

1. Inside the core, as for the solid cylinder: \(B = \dfrac{\mu_0Ir}{2\pi a^2}\).

2. In the gap the loop encloses the full core current: \(B = \dfrac{\mu_0I}{2\pi r}\).

3. Outside the shell the enclosed current is \(I - I = 0\), so \(B = 0\).

Current inside \(r\): \(I_{\text{enc}} = I\dfrac{r^2 - a^2}{b^2 - a^2} = 30 \times \dfrac{2.25 - 1}{4 - 1} = 12.5\) A. \(B = \dfrac{\mu_0I_{\text{enc}}}{2\pi r} = \dfrac{2 \times 10^{-7} \times 12.5}{0.015} \approx 1.7 \times 10^{-4}\) T.

Ampere's law is always true, but it gives \(B\) directly only for a loop with symmetry. On that loop \(B\) must have the same magnitude everywhere and point along the loop. A finite wire or a single loop lacks that symmetry, so \(\oint\vec B\cdot d\vec l\) cannot be written as \(B\) times a length. The Biot-Savart law must be used instead.

Take a rectangle with one side of length \(L\) along the axis inside and the opposite side far outside, where \(B \approx 0\). Its two short sides are perpendicular to \(\vec B\). Only the inside side contributes: \(\oint\vec B\cdot d\vec l = BL\). The rectangle is threaded by \(nL\) turns, each carrying \(I\): \(BL = \mu_0nLI\), so \(B = \mu_0nI\).

\(B = \mu_0nI = 4\pi \times 10^{-7} \times 1000 \times 2 \approx 2.5 \times 10^{-3}\) T.

1. \(n = \dfrac{2000}{0.5} = 4000\) turns per metre.

2. Only the side inside, along the axis: \(\vec B\) is along it. The outside side has \(B \approx 0\), and on the two short sides \(\vec B\) is perpendicular to \(d\vec l\).

3. \(B = \mu_0nI = 4\pi \times 10^{-7} \times 4000 \times 5 \approx 2.5 \times 10^{-2}\) T.

Deep inside, turns on both sides of a point contribute equally to \(\mu_0nI\). At an end, only the turns on one side exist, so the field is \(\dfrac{1}{2}\mu_0nI\). Two half-solenoids placed end to end would restore the full value.

Take a rectangular loop lying wholly outside. It encloses no current, so \(\oint\vec B\cdot d\vec l = 0\) for any such loop, which forces the outside field to be the same everywhere. Far away it is zero, so it is zero everywhere outside an ideal, infinitely long solenoid.

\(B = \mu_0nI\): double the current, or wind twice as many turns per metre. The radius does not appear, so a fatter solenoid with the same \(n\) and \(I\) has the same field inside.

\(B = \mu_r\mu_0nI = 500 \times 2.5 \times 10^{-3} \approx 1.26\) T.

Far outside on the axis the field is tiny. It grows as the end is approached and reaches \(\dfrac{1}{2}\mu_0nI\) at the end face. Inside it rises quickly and levels off at \(\mu_0nI\) over the middle part, where it is uniform. The pattern is symmetric about the centre.

1. The circle is threaded by all \(N\) turns: \(I_{\text{enc}} = NI = 500 \times 2 = 1000\) A turns.

2. \(B \times 2\pi r = \mu_0NI\), so \(B = \dfrac{\mu_0NI}{2\pi r} = \dfrac{4\pi \times 10^{-7} \times 1000}{2\pi \times 0.1} = 2 \times 10^{-3}\) T.

A circle in the hole threads no turns, so \(B = 0\) there. A circle outside the toroid is threaded by every turn twice, once going in and once coming out. The net current is zero and \(B = 0\). The field is confined to the core.

No. \(B = \dfrac{\mu_0NI}{2\pi r}\) depends on \(r\), so the field is stronger at the inner edge of the core than at the outer edge. For a thin core the variation is small.

If the core is thin, \(r\) is nearly the same across it and the turns per unit length along the core are \(n = \dfrac{N}{2\pi r}\). Then \(B = \dfrac{\mu_0NI}{2\pi r} = \mu_0nI\), the solenoid formula: a toroid is a solenoid bent into a ring.

\(n = \dfrac{400}{0.2} = 2000\) per metre. \(I = \dfrac{B}{\mu_0n} = \dfrac{5 \times 10^{-3}}{4\pi \times 10^{-7} \times 2000} \approx 2\) A.

By symmetry the field, if any, is circular about the axis with constant magnitude on a circle of radius \(r\). For \(r < R\) the circle encloses no current, so \(B \times 2\pi r = 0\) and \(B = 0\) everywhere inside the hollow.