4. Moving Charges And Magnetism

Magnetic interactions demand careful spatial thinking. Use the flat exam drawing to establish directions, then rotate the same arrangement in three dimensions to check which vectors are perpendicular.

Ørsted’s compass observation linked electric current with magnetism. Ampère and Biot–Savart developed quantitative descriptions of steady-current effects. The current SI now defines the ampere through the elementary charge; its older force-based definition remains historically useful.

A moving charge can interact with a magnetic field even when a stationary charge at the same position cannot.

Magnetism connects electric currents, forces and spatial directions. A compass near a current-carrying wire changes direction, showing that a current produces magnetic influence around it. The direction depends on current direction and geometry.

Electric and magnetic fields are different parts of the electromagnetic description. In this chapter the currents and fields are steady or slowly changing enough for magnetostatic formulas to be appropriate. Keep the source current, the observation point and the moving test charge distinct.

A charge is motionless in a region containing only a magnetic field. Does it initially experience magnetic force?

The magnetic part of the force contains the velocity vector.

Result: No. With v=0 the magnetic force is zero.

1. What simple observation links current with magnetic field?

Answers

1. A nearby compass changes orientation when current flows in a wire.

Magnetic force is perpendicular to both particle velocity and magnetic field.

The Lorentz force is $\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B)$. The magnetic part has magnitude $|q|vB\sin\theta$, where θ is the angle between v and B. Its SI unit defines the tesla through $1\ \mathrm T=1\ \mathrm{N\,s\,C^{-1}\,m^{-1}}$.

For a positive charge, curl the right hand from v towards B; the thumb gives force. Reverse the result for a negative charge. This is a cross product, so the order matters. A charge moving parallel or antiparallel to B has zero magnetic force.

For a straight wire segment in uniform B, $\mathbf F=I\mathbf L\times\mathbf B$, where L points along conventional current. Summing individual carrier forces gives this macroscopic result. For a bent wire or nonuniform field, integrate $d\mathbf F=I\,d\mathbf l\times\mathbf B$.

A +2 μC particle moves at 3000 m/s along +x in B=0.40 T along +y.

The vectors are perpendicular. $\hat x\times\hat y=\hat z$.

Result: 2.4 mN along +z.

A 0.25 m straight wire carries 4 A perpendicular to a 0.30 T field.

Use $F=ILB$.

Result: 0.3 N.

1. What magnetic force acts when θ=180°?

Answers

1. Zero.

A magnetic field bends a charged particle’s path without doing work on it.

Since $\mathbf F_B\perp\mathbf v$, its instantaneous power $\mathbf F_B\cdot\mathbf v$ is zero. In a magnetic field alone the speed and kinetic energy stay constant, while the direction may change.

For perpendicular entry into a uniform field, $|q|vB=mv^2/r$. Thus $r=mv/(|q|B)$ and $T=2\pi m/(|q|B)$. The angular frequency is $\omega=|q|B/m$. These are nonrelativistic results and ignore energy loss and collisions.

For oblique entry, resolve velocity into $v_\perp$ and $v_\parallel$. The perpendicular component gives circular motion of radius $mv_\perp/(|q|B)$; the parallel component is unchanged. Together they make a helix with pitch $v_\parallel T$.

A particle has m=4 × 10⁻²⁷ kg, |q|=2 × 10⁻¹⁹ C, v⊥=3 × 10⁵ m/s and B=0.20 T. Find radius.

Use the perpendicular speed in $r=mv_\perp/(|q|B)$.

Result: 30 mm.

A cyclotron uses this approximately speed-independent orbital period for nonrelativistic particles. An alternating electric field supplies energy across a gap while the magnetic field bends the path inside the electrodes. At relativistic speeds the simple frequency relation no longer keeps particles synchronised.

1. Does doubling nonrelativistic speed double the cyclotron period at fixed B?

Answers

1. No. It doubles radius, while period stays the same.

Each small current element contributes a field around it.

The Biot–Savart law in vacuum is $d\mathbf B=(\mu_0/4\pi)I\,d\mathbf l\times\hat{\mathbf r}/r^2$. The source element $d\mathbf l$ points with conventional current; r runs from the source element to the observation point. Contributions add as vectors. For numerical school calculations, $\mu_0\approx4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}$.

The current element is not a complete physical steady-current circuit by itself. The law is integrated over an actual closed current path or an appropriate long-wire model. A source element gives no contribution along its own line because the cross product vanishes there.

For a finite straight wire at perpendicular distance a, integration gives $B=\mu_0I(\sin\theta_1+\sin\theta_2)/(4\pi a)$ when the endpoint angles are taken from the perpendicular to each end on opposite sides of its foot. An infinitely long wire gives $B=\mu_0I/(2\pi a)$. Directions circle the wire by the right-hand grip rule.

A long straight wire carries 5 A. Find its field at 0.040 m.

Use $B=\mu_0I/(2\pi r)$; the π factors cancel.

Result: 25 μT.

1. At fixed current, what happens to the long-wire field when distance triples?

Answers

1. It becomes one third as large.

A circular loop’s symmetry cancels transverse field components on its axis.

For a loop of radius R carrying current I, every element is the same distance $\sqrt{R^2+x^2}$ from an axial point. Opposite elements cancel transverse components and their axial components add. Integrating around the circumference gives $B(x)=\mu_0IR^2/[2(R^2+x^2)^{3/2}]$. Multiply by N for N coincident turns.

At the centre, $B=\mu_0NI/(2R)$. Far along the axis, x≫R, $B\approx\mu_0NI R^2/(2x^3)$, the field of a magnetic dipole. Curl right-hand fingers with the current; the thumb gives the positive axial direction.

A 20-turn circular coil has radius 0.10 m and current 0.50 A. Find its central field.

Use $B=\mu_0NI/(2R)$ with the stated approximate μ₀.

Result: 62.8319 μT.

1. At fixed radius and current, what change follows from doubling the number of turns?

Answers

1. The axial field doubles.

Ampère’s circuital law connects magnetic circulation with linked current.

For steady currents in vacuum, $\oint\mathbf B\cdot d\mathbf l=\mu_0I_{\rm enc}$. The integration path is a closed mathematical curve, and the linked current crosses a spanning surface bounded by it. The positive current sense follows the right-hand rule for the chosen traversal direction.

The field on the path includes contributions from all currents. A current outside the spanning surface can alter B locally while contributing zero to its net circulation. As with Gauss’s law, symmetry is needed to extract a field magnitude from the integral.

For an infinitely long straight wire, choose a centred circle. B is tangential and has constant magnitude, so $B(2\pi r)=\mu_0I$. This reproduces the long-wire Biot–Savart result. For changing electric fields, the displacement-current term is needed; the simple magnetostatic law is not the full general law.

A circular path encloses currents +7 A and −2 A with respect to its normal. Find circulation using μ₀≈4π × 10⁻⁷ SI units.

Add signed linked currents before multiplying by μ₀.

Result: $62.8319\times10^{-7}\ \mathrm{T\,m}$.

1. Why is a circular path useful around a long straight wire?

Answers

1. Symmetry makes B tangential and constant on that circle.

A long tightly wound solenoid approximates a uniform magnetic field inside.

For a long solenoid with n turns per unit length carrying current I, an Ampèrian rectangle has one long side inside and one outside. Far from the ends, the outside field is small and inside field is nearly uniform. Then $Bl=\mu_0(nl)I$, giving $B\approx\mu_0nI$ in vacuum.

Finite solenoids have fringe fields near their ends; their outside field is not exactly zero. The direction follows the right-hand grip rule around the turns. Viewed from an end, counterclockwise current makes that end a north pole.

For a toroid, a circular path within the winding gives $B(r)=\mu_0NI/(2\pi r)$ in the ideal model. Its field varies with radial distance; it is approximately constant across a thin toroid. Ideal outside cancellation depends on the winding symmetry.

A 0.40 m solenoid has 800 turns and carries 0.30 A. Find its central field in the long-solenoid approximation.

First find n=N/L, then $B=\mu_0nI$.

Result: 0.754 mT.

1. If the same turns are spread over twice the length, what happens to the ideal internal field at fixed I?

Answers

1. It halves.

Parallel currents interact because each lies in the field made by the other.

For long parallel wires separated by d, wire 1 produces $B_1=\mu_0I_1/(2\pi d)$ at wire 2. The force on length L of wire 2 is $I_2LB_1$, so $F/L=\mu_0I_1I_2/(2\pi d)$. Parallel currents attract; antiparallel currents repel.

The equal-and-opposite interaction follows from the field directions and the wire-force rule. Increasing separation reduces the force per length. The approximation assumes long straight wires and neglects nearby return paths and end effects.

Historically the ampere was defined using the force between ideal parallel wires. In the present SI, the ampere is defined by fixing the elementary charge at $e=1.602176634\times10^{-19}\ \mathrm C$. Consequently μ₀ is measured rather than exactly fixed at $4\pi\times10^{-7}$; that value remains an excellent school-level approximation.

Two long wires carry 8 A and 3 A in the same direction, separated by 0.060 m. Find force per length.

Use $F/L=2\times10^{-7}I_1I_2/d$ with approximate μ₀.

Result: 80 μN/m, attractive.

1. What happens if one wire’s current reverses?

Answers

1. Attraction becomes repulsion, with the same magnitude if current magnitudes remain fixed.

A current loop behaves like a magnetic dipole in a uniform field.

For a rectangular loop, forces on opposite sides cancel in translation but can form a couple. Add their moments about the centre. The result does not depend on the particular rectangle dimensions once its area is fixed.

A coil has moment $\mathbf m=NI\mathbf A$, where the area vector follows the right-hand current rule. Its torque is $\boldsymbol\tau=\mathbf m\times\mathbf B$, magnitude $NIAB\sin\theta$. The angle is between the area normal and B, not between the plane of the loop and B.

Its potential energy is $U=-\mathbf m\cdot\mathbf B$ with zero at 90°. A uniform field gives no net force on the closed loop. A nonuniform field can produce a net force as well as torque. For a circular loop, $m=NI\pi R^2$ and the far axial field is $B=\mu_0(2m)/(4\pi x^3)$.

A 40-turn coil of area 0.0020 m² carries 0.30 A in B=0.50 T. Its normal is 30° from B. Find moment and torque.

Use m=NIA, followed by $\tau=mB\sin\theta$.

Result: m=0.024 A m²; τ=6 mN m.

1. Which alignment of a free magnetic dipole is stable?

Answers

1. Its moment parallel to B.

A moving-coil galvanometer converts current into a controlled angular deflection.

In a radial magnetic field, the coil’s magnetic torque has magnitude NIAB at every working angle. A spring supplies restoring torque κθ. At equilibrium $\kappa\theta=NIAB$, so $\theta/I=NAB/\kappa$. This ratio is current sensitivity. The radial arrangement makes deflection proportional to current.

If coil resistance is G, its voltage sensitivity is $\theta/V=NAB/(\kappa G)$. Increasing turns can improve current sensitivity but may also raise G, so it does not automatically improve voltage sensitivity by the same factor.

To make an ammeter of range I from a galvanometer with full-scale current I_g, connect a low shunt S in parallel: $I_gG=(I-I_g)S$, hence $S=I_gG/(I-I_g)$. To make a voltmeter of range V, add series resistance $R_s=V/I_g-G$. An ideal ammeter has very low resistance; an ideal voltmeter has very high resistance.

A galvanometer has G=100 Ω and full-scale current 1 mA. Find the shunt for a 1 A ammeter and the series resistor for a 5 V voltmeter.

Convert $I_g$ to amperes before substituting.

Result: S=0.1001 Ω; Rs=4900 Ω.

1. Why should a voltmeter have large resistance?

Answers

1. To draw little current and disturb the original circuit as little as possible.

Magnetic force is $q\mathbf v\times\mathbf B$ and does no work. Uniform B gives circular or helical motion in the nonrelativistic collision-free model.

Current sources produce B through Biot–Savart or Ampère’s law when their assumptions apply. A loop has moment NI A and torque $\mathbf m\times\mathbf B$. Galvanometers use this torque and a restoring spring.

• Always indicate the charge sign before applying a cross-product rule.
• A zero net force can coexist with nonzero torque.
• Field formulas require source geometry and approximation limits, not only matching units.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

4.1 A positive particle moves along +x while B points +z. Find magnetic force direction.

4.2 An electron has the same velocity and field as the preceding particle. Find its force direction.

4.3 A 5 μC charge moves at 2000 m/s perpendicular to B=0.20 T. Find force magnitude.

4.4 A wire of length 0.40 m carries 2 A at 30° to a 0.50 T field. Find force.

4.5 Why can a magnetic-only field not change a particle’s kinetic energy?

4.6 A nonrelativistic charged particle doubles its perpendicular speed at fixed B. State changes in radius and period.

4.7 A helical particle has vparallel=300 m/s and period 0.020 s. Find pitch.

4.8 A long wire carries 6 A. Find B at 0.030 m using approximate μ₀.

4.9 At what angle between a source current element and r does its Biot–Savart contribution vanish?

4.10 A 10-turn circular loop has R=0.050 m and I=0.40 A. Find central B.

4.11 Which components cancel at an axial point of a circular loop?

4.12 State what zero enclosed current tells you about an Ampèrian path.

4.13 A solenoid has 1500 turns/m and I=0.60 A. Estimate internal B.

4.14 Why does a finite solenoid depart from the ideal result near its ends?

4.15 A toroid has N=500, I=0.20 A and mean radius 0.10 m. Estimate B at that radius.

4.16 Two long parallel currents of 4 A and 5 A are 0.080 m apart. Find force per length.

4.17 How is the present SI ampere defined?

4.18 Is μ₀ exactly 4π × 10⁻⁷ N/A² in the present SI?

4.19 A 30-turn coil has area 0.004 m² and I=0.50 A. Find its moment.

4.20 The preceding coil’s normal is perpendicular to B=0.20 T. Find torque.

4.21 A loop lies with its plane perpendicular to B. What is its torque?

4.22 A galvanometer has Ig=2 mA and G=50 Ω. Find the shunt for a 0.50 A range.

4.23 For the same galvanometer, find the series resistance for a 10 V range.

4.24 Why does a radial field make galvanometer calibration linear?

4.25 Why does an ammeter connected across a cell risk damage?

4.26 At fixed B, charge and mass, what determines a helical path’s pitch?

Answers

4.1 −y.

4.2 +y.

4.3 2 mN.

4.4 0.2 N.

4.5 Its force is perpendicular to velocity, so it does no work.

4.6 Radius doubles; period is unchanged.

4.7 6 m.

4.8 40 μT.

4.9 0° or 180°.

4.10 50.2655 μT.

4.11 Components perpendicular to its axis.

4.12 Its magnetic circulation is zero; its field need not vanish everywhere.

4.13 1.131 mT.

4.14 Fringing fields and loss of translational symmetry become important.

4.15 0.2 mT.

4.16 50 μN/m; attraction for equal directions.

4.17 Through the fixed elementary charge e=1.602176634 × 10⁻¹⁹ C, with the second supplying the time unit.

4.18 No; that is an extremely close approximation.

4.19 0.06 A m².

4.20 12 mN m.

4.21 Zero; its normal is parallel or antiparallel to B.

4.22 0.2008 Ω.

4.23 4950 Ω.

4.24 The magnetic torque is NIAB independent of deflection, while spring torque is proportional to angle.

4.25 Its low resistance can draw excessive current, effectively approaching a short circuit.

4.26 Its velocity component parallel to B multiplied by its cyclotron period.