Sources of Magnetic Field

Current out of the page: field lines circle counterclockwise as seen from the front (thumb toward you, fingers curl counterclockwise).

• (a) At the point to the right, counterclockwise means \(+y\) (up). Check with Biot–Savart: \(d\boldsymbol{\ell}\times\hat{\mathbf{r}} = \hat{\mathbf{z}}\times\hat{\mathbf{x}} = +\hat{\mathbf{y}}\). ✓
• (b) At the point above, counterclockwise means \(-x\) (left). Check: \(\hat{\mathbf{z}}\times\hat{\mathbf{y}} = -\hat{\mathbf{x}}\). ✓
• (c) Fingers clockwise in the page: the thumb points into the page.
• (d) Never. \(\mathbf{B}\) of a straight wire is always tangent to circles around it — perpendicular to the radial direction.

(a) At \(x = 5.0\) cm: wire 1 (out of page, counterclockwise field) gives \(+y\) at a point to its right. Wire 2 (into page, clockwise field) gives \(+y\) at a point to its left. They add:

\(B = 64\ \mu\text{T}\) in the \(+y\) direction.

(b) Between the wires the fields add, so the zero must be outside, nearer the smaller current: to the right of wire 2, at \(x > 0.10\) m. There wire 1 gives \(+y\) and wire 2 gives \(-y\). Setting magnitudes equal:

\(x = 25\) cm (15 cm beyond wire 2). Check: \(10/0.25 = 40\) and \(6.0/0.15 = 40\). ✓

(a) Each element of the arc is a distance \(R\) from the centre and perpendicular to \(\hat{\mathbf{r}}\) (\(\sin\theta = 1\)), and all contributions point the same way (along the arc's axis). With \(d\ell = R\,d\phi'\):

(b) Three pieces, and by the right-hand rule all three give a field in the same direction at the centre (perpendicular to the plane of the hairpin):

• Semicircle (\(\phi = \pi\)): \(\dfrac{\mu_0 I}{4R}\).
• Each straight section is semi-infinite and the centre is opposite its end, at perpendicular distance \(R\): \(\dfrac{\mu_0 I}{4\pi R}\) each.

\(B\approx 0.129\) mT, perpendicular to the plane. The straight parts contribute \(5.0\times10^{-5}\) T together — not negligible.

Answers: 25.1 μT, 8.89 μT, 7.66 cm. The field falls off quickly along the axis — within less than one radius it has halved.

Opposite currents: repulsive. The current rose by a factor 20, so the force rose by \(20^2 = 400\) (\(16\times400 = 6400\)). ✓ That \(I^2\) scaling is why fault ratings, not normal ratings, set the mechanical design of switchgear.

(a) Add up the current in thin rings of radius \(r'\), width \(dr'\), area \(2\pi r'\,dr'\):

(b) By symmetry \(\mathbf{B}\) circles the axis with magnitude depending only on \(r\). Amperian circle of radius \(r\): \(\oint\mathbf{B}\cdot d\boldsymbol{\ell} = B(2\pi r)\).

(c) At \(r = R\) the inside formula gives \(\mu_0 I R^2/(2\pi R^3) = \mu_0 I/(2\pi R)\), the same as the outside one. ✓ Compare a uniform current, where \(B\propto r\) inside: here more current sits near the surface, so \(B\propto r^2\) — smaller deep inside.

(a) 0.48 mT, (b) 0.60 mT, (c) zero. In (c), Ampère gives \(B(2\pi r) = 0\), and symmetry says \(B\) is the same all round, so \(B\) itself vanishes.

(d) The magnetic field (and, similarly, the electric field) of the signal is confined between the conductors. The cable neither radiates into nearby circuits nor, by the reverse argument, picks up their fields — which is why coax carries RF, video and test-equipment signals.

(a) 8.0 mT; (b) about 600 A — a job for superconducting wire.

(c) The rectangular Amperian loop has one side of length \(\ell\) inside, parallel to \(\mathbf{B}\); the other sides contribute nothing. So \(B\ell = \mu_0(n\ell)I\). Only the turns per length and the current enter — the cross-section never appears.

(a) 3.0 mT, (b) 2.0 mT. (c) In the hole, an Amperian circle encloses no current and by symmetry \(B = 0\). (d) Outside, the circle is threaded by \(N\) wires going one way and \(N\) coming back: \(B = 0\).

The field is 50% stronger at the inner edge than the outer — a toroid is not uniform, unlike an ideal solenoid.

(a) Copper: diamagnetic, very weakly repelled. Aluminium: paramagnetic, very weakly attracted. Iron: ferromagnetic, strongly attracted.

Silicon steel saturates at roughly 1.5–2 T: once all the domains are aligned, \(\mu_r\) is no longer constant and the field cannot follow the formula. The core would be deep in saturation, its effective permeability would collapse, and an inductor built this way would lose most of its inductance.

(c) Reduce \(NI\) (fewer turns or less current) so \(B\) stays below about 1 T, or use a larger core (bigger \(r\)), or introduce an air gap in the core, which lowers the effective permeability and lets the core handle more ampere-turns before saturating — standard practice in power inductors.