Magnetic Effects of Current and Magnetism • Topic 2 of 3

Force on Charges & Conductors

The Lorentz force. A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a magnetic force $\vec{F}=q(\vec{v}\times\vec{B})$. Its magnitude is $F=qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. The force is maximum when $\theta=90^\circ$ and zero when the charge moves parallel to the field. If an electric field is also present, the total Lorentz force is $\vec{F}=q\vec{E}+q(\vec{v}\times\vec{B})$.

  • The magnetic force is always perpendicular to $\vec{v}$, so it does no work on the charge. It changes the direction of motion, not the speed (and hence not the kinetic energy).
  • The direction is given by the right-hand rule for $\vec{v}\times\vec{B}$ (reversed for a negative charge).

Circular motion of a charge. When a charge enters a uniform field perpendicularly, the constant perpendicular force makes it move in a circle. The magnetic force supplies the centripetal force: $qvB=\frac{mv^2}{r}$, giving the radius $r=\frac{mv}{qB}$. The time period $T=\frac{2\pi m}{qB}$ is independent of speed and radius, and the frequency $f=\frac{qB}{2\pi m}$ is the cyclotron frequency.

The cyclotron. A cyclotron accelerates positive ions to high energies. Two hollow D-shaped electrodes (dees) sit in a strong magnetic field; a high-frequency alternating voltage across the gap accelerates the ion each time it crosses. Inside the dees the field bends it in a semicircle. As speed rises the radius grows, but the period stays the same (resonance condition $f=\frac{qB}{2\pi m}$). The maximum energy is $E_{max}=\frac{q^2B^2R^2}{2m}$, where $R$ is the dee radius.

Force on a current-carrying conductor. A straight wire of length $L$ carrying current $I$ in a field $\vec{B}$ feels a force $\vec{F}=I\vec{L}\times\vec{B}$, of magnitude $F=BIL\sin\theta$. The direction is given by Fleming's left-hand rule (forefinger = field, middle finger = current, thumb = force).

Force between parallel currents. Two long parallel wires a distance $d$ apart carrying currents $I_1$ and $I_2$ exert a force per unit length $\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}$ on each other. Like (parallel) currents attract; opposite (antiparallel) currents repel. This relation defines the ampere: equal currents of 1 A in wires 1 m apart exert $2\times10^{-7}$ N/m.

Torque on a current loop. A rectangular coil of $N$ turns, area $A$, carrying current $I$ in a field $B$ experiences a torque $\tau=NIAB\sin\theta$, where $\theta$ is the angle between the field and the normal to the coil. Defining the magnetic moment $\vec{m}=NI\vec{A}$, this becomes $\vec{\tau}=\vec{m}\times\vec{B}$. The torque is maximum when the coil's plane is parallel to $\vec{B}$ ($\theta=90^\circ$) and zero when the plane is perpendicular to it.

Moving-coil galvanometer. A galvanometer detects small currents. A coil pivoted in a radial magnetic field turns until the deflecting torque $NIAB$ balances the restoring torque $k\phi$ of a spring, so the deflection $\phi=\frac{NAB}{k}I$ is proportional to the current. A radial field keeps the scale linear. The galvanometer becomes an ammeter with a low-resistance shunt in parallel, and a voltmeter with a high resistance in series.

Charged particle in a uniform magnetic field: circular path produced by the Lorentz force F equals q v cross BB into pagex x xx x x+qvF = q v x Bradius r = mv / qB
Solved Examples
1
Worked Example
A proton moves with a speed of $2\times10^6$ m/s perpendicular to a magnetic field of 0.5 T. Find the magnetic force on it. (charge $=1.6\times10^{-19}$ C)
Solution
  1. Step 1: Use $F=qvB\sin\theta$ with $\theta=90^\circ$, so $\sin\theta=1$.
  2. Step 2: $F=qvB=(1.6\times10^{-19})(2\times10^6)(0.5)$.
  3. Step 3: $F=1.6\times10^{-19}\times1\times10^6=1.6\times10^{-13}$ N.

Answer: $F=1.6\times10^{-13}\,\text{N}$, perpendicular to both $\vec{v}$ and $\vec{B}$.

2
Worked Example
An electron (mass $9.1\times10^{-31}$ kg, charge $1.6\times10^{-19}$ C) moves at $3\times10^7$ m/s perpendicular to a field of 0.2 T. Find the radius of its circular path.
Solution
  1. Step 1: The magnetic force is centripetal: $r=\frac{mv}{qB}$.
  2. Step 2: Substitute: $r=\frac{(9.1\times10^{-31})(3\times10^7)}{(1.6\times10^{-19})(0.2)}$.
  3. Step 3: Numerator $=2.73\times10^{-23}$; denominator $=3.2\times10^{-20}$.
  4. Step 4: $r=\frac{2.73\times10^{-23}}{3.2\times10^{-20}}=8.5\times10^{-4}$ m.

Answer: $r\approx8.5\times10^{-4}\,\text{m}=0.85\,\text{mm}$.

3
Worked Example
A straight wire of length 0.20 m carries a current of 4 A at right angles to a uniform field of 0.30 T. Find the force on the wire.
Solution
  1. Step 1: Use $F=BIL\sin\theta$ with $\theta=90^\circ$.
  2. Step 2: $F=(0.30)(4)(0.20)(1)$.
  3. Step 3: $F=0.24$ N.
  4. Step 4: Direction by Fleming's left-hand rule, perpendicular to both wire and field.

Answer: $F=0.24\,\text{N}$, perpendicular to the wire and the field.

4
Worked Example
Two long parallel wires 0.05 m apart carry currents of 10 A and 15 A in the same direction. Find the force per unit length and state its nature.
Solution
  1. Step 1: Force per unit length $\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}$.
  2. Step 2: Substitute: $\frac{F}{L}=\frac{(4\pi\times10^{-7})(10)(15)}{2\pi(0.05)}$.
  3. Step 3: $=\frac{2\times10^{-7}\times150}{0.05}=\frac{3\times10^{-5}}{0.05}=6\times10^{-4}$ N/m.
  4. Step 4: Currents are in the same direction, so the wires attract.

Answer: $\frac{F}{L}=6\times10^{-4}\,\text{N/m}$, attractive (parallel currents attract).

5
Worked Example
A coil of 50 turns and area $2\times10^{-3}\,\text{m}^2$ carries 3 A in a field of 0.4 T. Find the maximum torque on it.
Solution
  1. Step 1: Torque $\tau=NIAB\sin\theta$; maximum when $\sin\theta=1$.
  2. Step 2: $\tau_{max}=NIAB=(50)(3)(2\times10^{-3})(0.4)$.
  3. Step 3: $=50\times3\times2\times10^{-3}\times0.4=0.12$ N m.

Answer: $\tau_{max}=0.12\,\text{N m}$ (when the coil's plane is parallel to the field).

6
Worked Example
Why does the magnetic force do no work on a moving charge, and what is the role of a radial field in a moving-coil galvanometer?
Solution
  1. Step 1: The magnetic force $\vec{F}=q(\vec{v}\times\vec{B})$ is always perpendicular to $\vec{v}$.
  2. Step 2: Work $=\vec{F}\cdot\vec{v}=0$ since the force has no component along the motion; speed and kinetic energy stay constant.
  3. Step 3: In a galvanometer a radial field makes the plane of the coil always parallel to $\vec{B}$, so $\sin\theta=1$ for every position.
  4. Step 4: The deflecting torque is then $NIAB$, independent of $\phi$, giving $\phi\propto I$ — a linear scale.

Answer: The force is perpendicular to velocity so it does no work; a radial field keeps $\sin\theta=1$, making the galvanometer's deflection proportional to current.

Key Points

  • Lorentz force: $\vec{F}=q\vec{E}+q(\vec{v}\times\vec{B})$; the magnetic part $F=qvB\sin\theta$ does no work (perpendicular to $\vec{v}$).
  • A charge in a perpendicular field moves in a circle of radius $r=\frac{mv}{qB}$ with period $T=\frac{2\pi m}{qB}$ (cyclotron frequency $f=\frac{qB}{2\pi m}$).
  • Force on a current wire: $\vec{F}=I\vec{L}\times\vec{B}$, magnitude $BIL\sin\theta$ (Fleming's left-hand rule).
  • Parallel currents: $\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}$ — like currents attract, unlike repel (defines the ampere).
  • Torque on a loop: $\tau=NIAB\sin\theta=\vec{m}\times\vec{B}$; a moving-coil galvanometer uses a radial field so $\phi\propto I$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The magnetic force on a charge moving parallel to the field is:
Explanation: $F=qvB\sin\theta$; for $\theta=0^\circ$, $\sin\theta=0$, so the force is zero.
Q2.The radius of the circular path of a charged particle in a magnetic field is:
Explanation: $qvB=\frac{mv^2}{r}\Rightarrow r=\frac{mv}{qB}$.
Q3.Two parallel wires carrying current in opposite directions will:
Explanation: Antiparallel (opposite) currents repel; parallel currents attract.
Q4.The torque on a current loop of $N$ turns in a field $B$ is:
Explanation: $\tau=NIAB\sin\theta=\vec{m}\times\vec{B}$ where $m=NIA$.
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