20. Magnetic fields

Separate field source from test object: stationary charges do not feel magnetic force, while currents and moving charges can.

A current in a wire creates a field around the wire; a nearby compass aligns with the local field direction.

Lines form continuous loops, are denser where the field is stronger and do not cross at a point.

Around a long straight current-carrying wire, field lines are concentric circles whose direction follows the right-hand grip rule.

Reverse current or field direction to reverse force. Parallel current and field produce no force in the ideal straight-wire model.

A wire between magnet poles deflects when current flows, forming the basis of a simple motor.

At θ=90°, F=BIL is maximum. At θ=0° or 180°, F=0. Reversing either I or B reverses force; reversing both leaves force direction unchanged.

A 0.20 m wire carrying 3.0 A at 30° to a 0.50 T field experiences F=(0.50)(3.0)(0.20)sin30°=0.150 N. Use the left hand separately to give direction.

The definition assumes the conductor is perpendicular; otherwise divide by IL sinθ or resolve the perpendicular component.

A 0.40 N force on a 0.20 m wire carrying 2.0 A gives B=1.0 T when perpendicular.

On diagrams, × means into the page and • means out of the page. Reversing v, B or charge sign reverses force; reversing any two leaves force direction unchanged.

Because force is perpendicular to velocity, it does no work on an isolated particle: speed and kinetic energy stay constant while direction changes.

Force is maximum, B|Q|v, for perpendicular motion. It is zero for v parallel/antiparallel to B and for a stationary particle.

A proton moving at 2.0×10⁶ m s⁻¹ perpendicular to B=0.30 T experiences F=(0.30)(1.60×10⁻¹⁹)(2.0×10⁶)=9.6×10⁻¹⁴ N.

Let current I flow along the conductor, B be perpendicular to the broad face, width across the Hall contacts be w, and thickness parallel to B be t. Carrier number density is n, carrier charge magnitude q and drift speed v.

For B=4.0×10⁻⁶ T, I=5.4 A, n=1.5×10¹⁶ m⁻³, t=1.8×10⁻³ m and q=1.60×10⁻¹⁹ C, V_H=BI/(ntq)=5.0 V.

A semiconductor such as silicon has much smaller n than copper, so it produces a larger, easier-to-measure Hall voltage for the same B,I,t and q.

With sensitivity 20 mV T⁻¹, a maximum reading of +6.0 mV gives B=6.0/20=+0.30 T along the calibrated positive normal.

If the field occupies only part of space, the particle follows a circular arc inside and then continues in a straight line tangent to the arc after leaving the field.

This condition assumes perpendicular, uniform fields and the correct orientation; faster or slower particles deflect.

E=2.0×10⁴ N C⁻¹ and B=0.50 T select v=4.0×10⁴ m s⁻¹.

For a straight wire, point the right thumb along conventional current; curled fingers give circular field direction. Current into the page gives clockwise arrows; current out gives anticlockwise arrows.

For a circular coil or solenoid, curl right-hand fingers with conventional current around the turns; the thumb points through the coil toward its magnetic north end and gives the internal field direction.

Reversing current reverses every field arrow and swaps the coil/solenoid north and south poles, while leaving the field-line shape unchanged.

Use the right-hand grip rule for polarity and distinguish ideal uniform interior field from fringing at the ends.

Increasing turns per metre or current strengthens the field, while reversing current reverses the poles.

Use the field of one wire and F=BIL on the other; the force per length falls as separation increases.

Two long parallel wires carrying equal currents in the same direction pull toward each other.

Use the component of B perpendicular to the area and state the surface orientation.

A 0.20 T field through 0.50 m² perpendicular surface gives 0.10 Wb.

For B=7.2×10⁻³ T through A=3.2×10⁻⁴ m² normally, Φ=(7.2×10⁻³)(3.2×10⁻⁴)=2.3×10⁻⁶ Wb.

At fixed orientation, doubling B or A doubles Φ. For a circular loop use A=πr² and convert all lengths to metres before squaring.

Changing turns, field, area or orientation can change linkage. Distinguish flux through one turn from linkage of the whole coil.

A 100-turn coil with 2 mWb per turn has flux linkage 0.20 Wb-turn.

Place a primary coil connected to a switchable supply beside a secondary coil connected to a galvanometer. Closing or opening the switch gives only a transient secondary deflection; steady primary current gives none. Reversing primary current reverses the induced deflection.

An induced emf exists when linkage changes even if the circuit is open; an induced current requires a complete conducting path.

If flux per turn of a 200-turn coil changes from 3.0 μWb to 0.50 μWb in 0.010 s, |ε|=200(2.5×10⁻⁶)/0.010=0.050 V.

Opposition is required by energy conservation: if induced effects assisted the change, the system could amplify motion and electrical output without external work.