Magnetism

Like poles repel and unlike poles attract. Two north poles are alike, so they push apart.

By convention, field lines leave the north pole and curve around to enter the south pole outside the magnet (and run S to N inside it), forming continuous closed loops.

Iron, nickel and cobalt are the common ferromagnetic materials – they are strongly attracted to magnets and can themselves be magnetised. Copper and aluminium are metals but are not ferromagnetic.

The spacing of field lines shows the field strength: closely spaced lines (as at A) mean a strong field, widely spaced lines (as at B) a weak one. That is why the lines crowd together near a magnet's poles.

The Earth has its own magnetic field, as though a huge bar magnet sat inside it. A compass needle is a small magnet that lines up with that field, so one end points toward the Earth's magnetic north.

The field circles the wire in closed loops – concentric circles in the plane perpendicular to the wire. Their direction is given by the right-hand grip rule, and they get weaker farther from the wire.

$B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi\times10^{-7})(10)}{2\pi(0.050)} = \dfrac{4\times10^{-6}}{0.10} = 4\times10^{-5}\text{ T}$.

Oersted's experiment linked electricity and magnetism: a moving charge – an electric current – sets up a magnetic field in the space around it. This is the basis of every electromagnet and of the field $B = \dfrac{\mu_0 I}{2\pi r}$ that circles a wire.

Around a straight wire $B \propto \dfrac{1}{r}$ (an inverse first-power law, not inverse-square). Doubling the distance therefore halves the field.

Point the right thumb out of the page (the current direction); the fingers curl anticlockwise. So the field circles the wire anticlockwise. (For current into the page it would be clockwise.)

$B = \mu_0 n I = (4\pi\times10^{-7})(1000)(2.0) = 8\pi\times10^{-4} \approx 2.5\times10^{-3}\text{ T}$.

Grip the wire with the right hand, thumb pointing east (the current). The curled fingers point north on the underside of the wire, so the field directly below points north. (Directly above the wire it would point south, and reversing the current flips both.)

Curl the right hand's fingers the way the current flows (clockwise as viewed from the right); the thumb then points into the solenoid, toward the left. The thumb gives the north-pole direction, so the field inside points left and emerges at the left end – the north pole. (Viewed end-on, a clockwise current makes that end a south pole.)

The field of a coil grows with the current and with the number of turns ($B = \mu_0 n I$), and an iron core boosts it further. So reducing the current weakens the electromagnet; all the other options strengthen it.

A ferromagnetic material is made of tiny regions called domains, each already magnetised. Unmagnetised, they point every which way and cancel out. Magnetising the material lines the domains up, so their fields add and the whole piece becomes a magnet.

The Earth's magnetic poles sit near, but not exactly at, the geographic poles, and they slowly drift. So a compass points close to true north but usually a little off – the difference is called magnetic declination.

Rearranging, $I = \dfrac{2\pi r B}{\mu_0} = \dfrac{2\pi (0.10)(3\times10^{-5})}{4\pi\times10^{-7}} = \dfrac{0.6\pi\times10^{-5}}{4\pi\times10^{-7}} = 15\text{ A}$.

The field inside a long solenoid, $B = \mu_0 n I$, depends only on the turns per metre and the current – there is no diameter or area term. So making it wider does not change the field inside.

Since $B = \dfrac{\mu_0 I}{2\pi r}$, doubling $I$ multiplies $B$ by $2$, and halving $r$ multiplies it by another $2$: $2 \times 2 = 4$ times as strong.

From $B = \mu_0 n I$, $n = \dfrac{B}{\mu_0 I} = \dfrac{5.0\times10^{-3}}{(4\pi\times10^{-7})(2.0)} \approx 2000\text{ turns/m}$. Over $0.10\text{ m}$ that requires $N = nL = 2000 \times 0.10 = 200$ turns.

First $n = \dfrac{N}{L} = \dfrac{1000}{0.20} = 5000\text{ turns/m}$. Then $B = \mu_0 n I = (4\pi\times10^{-7})(5000)(3.0) = 6\pi\times10^{-3} \approx 1.9\times10^{-2}\text{ T}$.

Soft iron is magnetically soft: its domains line up readily to give a strong field, then fall back out of alignment once the current stops. That lets the electromagnet be switched on and off. Hard steel would stay magnetised, making a permanent magnet instead.

By the right-hand rule, at the midpoint wire 1's field points one way (say up) and wire 2's field points the opposite way (down). Being equal in size and opposite in direction, they cancel exactly, so the field at P is zero. (If the currents were in opposite directions, the two fields would instead add.)

Since $B = \mu_0 n I$ and $n = \dfrac{N}{L}$, at fixed length and current the field is proportional to the number of turns $N$. Doubling the turns doubles $n$, and so doubles $B$. (Diameter does not appear in the formula.)

$B = \mu_0 n I$. Doubling the turns (same length) doubles $n$, multiplying the field by $2$; halving the current multiplies it by $\tfrac{1}{2}$. Together $2 \times \tfrac{1}{2} = 1$, so the field is unchanged.