JEE Physics Moving Charges and Magnetism Notes Set 03

Magnetic field and the Biot-Savart law

A moving charge, or equivalently a current, sets up a magnetic field in the space around it. The field strength is measured by the magnetic induction \( \vec B \), whose SI unit is the tesla (1 T = 1 Wb/m² = 1 N/A m) and whose dimensions are \( [MT^{-2}A^{-1}] \).
The contribution of a short current element \( I\,d\vec l \) to the field at a point with position vector \( \vec r \) from the element is given by the Biot-Savart law:
\[ d\vec B=\frac{\mu_0}{4\pi}\,\frac{I\,d\vec l\times\vec r}{r^3},\qquad dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2},\qquad \frac{\mu_0}{4\pi}=10^{-7}\ \text{T m/A} \]
The field is zero along the line of the element (theta = 0 or 180°), is largest at right angles to it, and its direction is perpendicular to the plane containing \( d\vec l \) and \( \vec r \) (use the right-hand screw rule). A point charge q moving with velocity v produces \( \vec B=\frac{\mu_0}{4\pi}\frac{q\,\vec v\times\vec r}{r^3} \), because \( I\,d\vec l \) is the same as \( q\vec v \).

Standard results from Biot-Savart

1. Straight wire of finite length, point at perpendicular distance d, with the lines to the two ends making angles alpha and beta with the perpendicular:
\[ B=\frac{\mu_0 I}{4\pi d}\,(\sin\alpha+\sin\beta) \]
2. Infinitely long wire: alpha = beta = 90°, so \( B=\frac{\mu_0I}{2\pi d} \). Near the end of a semi-infinite wire (perpendicular foot at the end): \( B=\frac{\mu_0I}{4\pi d} \). For a point on the line of the wire, B = 0. The field lines are circles about the wire, as in the figure.

3. Circular arc of radius R and angle \( \theta \) (radians), at the centre: \( B=\frac{\mu_0I\theta}{4\pi R} \). Full loop: \( \frac{\mu_0I}{2R} \); semicircle: \( \frac{\mu_0I}{4R} \); a coil of N turns gives N times the loop value.
4. On the axis of a circular loop at distance x from its centre: \( B=\frac{\mu_0 I R^2 N}{2(R^2+x^2)^{3/2}} \). Far away (x much larger than R) this becomes \( \frac{\mu_0}{4\pi}\frac{2m}{x^3} \) with m = NIA, the same form as the field of an electric dipole.
5. Solenoid with n turns per unit length: on the axis, \( B=\frac{\mu_0nI}{2}(\sin\alpha+\sin\beta) \). Deep inside a long solenoid \( B=\mu_0nI \), and at the centre of one end it is half of that, \( \tfrac12\mu_0nI \).
When a conductor is made of several straight and curved pieces, find each piece's contribution separately, giving a sign for the direction (into or out of the page), and add. Pieces that point straight at the observation point contribute nothing.

Ampere's circuital law

The line integral of the magnetic field around any closed path equals \( \mu_0 \) times the net current passing through the surface bounded by the path:
\[ \oint\vec B\cdot d\vec l=\mu_0 I_{enclosed} \]
The sign of each current is fixed by the right-hand rule applied to the sense in which the loop is traversed, and currents outside the loop do not enter the right-hand side (they do affect B itself). The law is most useful when symmetry lets B be taken out of the integral.
1. Long straight wire: \( B=\frac{\mu_0I}{2\pi r} \).
2. Solid cylinder of radius a with uniform current: for r less than a, \( B=\frac{\mu_0Ir}{2\pi a^2} \) (grows linearly); for r greater than a, \( B=\frac{\mu_0I}{2\pi r} \). For a thin hollow tube, B = 0 inside and the same outside formula applies.
3. Long solenoid: \( B=\mu_0nI \) inside and nearly zero outside. Toroid with N turns and mean radius r: \( B=\frac{\mu_0NI}{2\pi r} \) inside the winding and zero outside.
4. Coaxial cable with equal and opposite currents: the field is zero outside because the enclosed current is zero.

Force on a moving charge

A charge q moving with velocity v in a magnetic field B feels \( \vec F=q\,\vec v\times\vec B \), of magnitude \( qvB\sin\theta \). The force is perpendicular to both v and B, so it never changes the speed or kinetic energy of the charge. In the presence of an electric field too, the Lorentz force is \( \vec F=q(\vec E+\vec v\times\vec B) \). Unlike the electric force, the magnetic force acts only on moving charges and does no work.
In a region with crossed fields E and B, a charge passes undeflected if \( v=E/B \) (the principle of the velocity selector).

Motion in a uniform magnetic field

1. Velocity parallel to B: no force, uniform straight-line motion.
2. Velocity perpendicular to B: the magnetic force provides the centripetal force, so the path is a circle (figure):
\[ r=\frac{mv}{qB}=\frac{p}{qB}=\frac{\sqrt{2mK}}{qB},\qquad T=\frac{2\pi m}{qB},\qquad f=\frac{qB}{2\pi m} \]
The period does not depend on the speed or the radius. If a charge is accelerated through a potential difference V, then \( r=\frac1B\sqrt{\frac{2mV}{q}} \).

3. Velocity at an angle \( \theta \) to B: the component \( v\cos\theta \) along B stays constant and the component \( v\sin\theta \) gives circular motion. The path is a helix:
\[ r=\frac{mv\sin\theta}{qB},\qquad T=\frac{2\pi m}{qB},\qquad \text{pitch}=v\cos\theta\cdot T=\frac{2\pi mv\cos\theta}{qB} \]

If an electric field is also present along B, the circular part is unchanged but the axial speed grows steadily, so the pitch increases with time.
The cyclotron uses this idea: a charge circles inside two hollow dees, and an alternating voltage of frequency \( f=\frac{qB}{2\pi m} \) gives it an energy gain of qV at each crossing of the gap, i.e. 2qV per revolution. The radius of the orbit grows with the square root of the energy.

Force on a current-carrying conductor

A current element feels \( d\vec F=I\,d\vec l\times\vec B \). For a straight wire of length l in a uniform field, \( F=BIl\sin\theta \). A wire of any shape between two fixed points feels the same force in a uniform field as a straight wire joining those points, and a closed loop in a uniform field feels zero net force.
Two long parallel wires with currents \( I_1 \) and \( I_2 \) at separation d exert on each other, per unit length,
\[ \frac Fl=\frac{\mu_0I_1I_2}{2\pi d} \]
Currents in the same direction attract and opposite currents repel. This is the basis of the definition of the ampere.

Current loop as a magnetic dipole

A loop of area A and current I (N turns) has magnetic moment \( \vec m=NI\vec A \), directed along the normal given by the right-hand screw rule; unit A m². A charge q circling with speed v in a circle of radius r has \( m=\frac{qvr}{2}=\frac{q}{2m_0}L \) (with L the angular momentum and m0 the particle mass), and for an electron the moment points opposite to L.
In a uniform field B the net force is zero, but there is a torque and a potential energy:
\[ \vec\tau=\vec m\times\vec B\ (\tau=mB\sin\theta),\qquad U=-\vec m\cdot\vec B=-mB\cos\theta,\qquad W=mB(\cos\theta_1-\cos\theta_2) \]
Stable equilibrium is at theta = 0 (U = -mB), unstable equilibrium at 180° (U = +mB), and the torque is largest (mB) at 90°. The work in turning from parallel to antiparallel is 2mB. In a non-uniform field a dipole also feels a net force along the field gradient. On the axis of the loop, at a large distance x, \( B=\frac{\mu_0}{4\pi}\frac{2m}{x^3} \); the formulas for a magnetic dipole mirror those for an electric dipole with \( \frac1{4\pi\varepsilon_0}\to\frac{\mu_0}{4\pi} \).

Moving coil galvanometer

A coil of N turns and area A is suspended between cylindrical pole pieces with a soft-iron core, making the field radial so that the coil's plane is always parallel to B. The deflecting torque NIAB is balanced by the restoring torque of the suspension, \( k\phi \):
\[ NIAB=k\phi\ \Rightarrow\ I=\frac{k}{NAB}\,\phi \]
Hence the deflection is proportional to the current and the scale is linear. Current sensitivity is \( \phi/I=NAB/k \) and voltage sensitivity is \( \phi/V=NAB/(kR) \). A shunt in parallel turns the galvanometer into an ammeter, and a large series resistance turns it into a voltmeter.

Worked examples

Illustration 1. Find B at the centre O of a long wire bent into a semicircle of radius R with two long straight parts leaving tangentially, the current being i.
Answer:
Each long straight piece ends at the foot of the perpendicular from O (distance R), so it contributes \( \frac{\mu_0i}{4\pi R} \) and the semicircle gives \( \frac{\mu_0i}{4R} \). All three point the same way.
\( B=\frac{\mu_0i}{4\pi R}+\frac{\mu_0i}{4\pi R}+\frac{\mu_0i}{4R}=\frac{\mu_0i}{4\pi R}(\pi+2) \).

Illustration 2. A charge of 1 C sits at one end of a non-conducting rod of length 0.6 m which rotates in a vertical plane about a horizontal axis through its other end, at angular speed \( 10^4\pi \) rad/s. Find B on the axis of rotation at a distance 0.8 m from the centre of the path.
Answer:
Equivalent current \( i=qf=1\times\frac{10^4\pi}{2\pi}=5\times10^3\ \text{A} \).
On the axis of the circular path: \( B=\frac{\mu_0}{4\pi}\frac{2\pi iR^2}{(R^2+x^2)^{3/2}}=10^{-7}\times\frac{2\pi\times5\times10^3\times0.36}{(0.36+0.64)^{3/2}} \).
\( B\approx1.13\times10^{-3}\ \text{T} \).

Illustration 3. An electron has a kinetic energy of 1 keV and moves perpendicular to a uniform field B = 10^-4 T. Find the period and the radius.
Answer:
\( T=\frac{2\pi m}{eB}=\frac{2\pi\times9.1\times10^{-31}}{1.6\times10^{-19}\times10^{-4}}=3.6\times10^{-7}\ \text{s} \).
\( p=\sqrt{2mK}=\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-16}}=1.7\times10^{-23}\ \text{kg m/s} \).
\( r=\frac{p}{eB}=\frac{1.7\times10^{-23}}{1.6\times10^{-19}\times10^{-4}}\approx1.07\ \text{m} \).

Illustration 4. An alpha particle (mass \( 6.4\times10^{-27} \) kg, charge 2e) is accelerated through \( 10^4 \) V and enters normally a field region of thickness 0.1 m with B = 0.1 T. Find the angle through which its direction changes.
Answer:
Kinetic energy \( K=2e\times10^4=3.2\times10^{-15}\ \text{J} \), so \( r=\frac{\sqrt{2mK}}{qB}=\frac{\sqrt{2\times6.4\times10^{-27}\times3.2\times10^{-15}}}{3.2\times10^{-19}\times0.1}=0.2\ \text{m} \).
For a region of thickness d, \( \sin\theta=\frac dr=0.5 \), so the deviation is \( 30^\circ \).

Illustration 5. A beam of protons with speed \( 4\times10^5 \) m/s enters a uniform field of 0.3 T at 60° to the field. Find the radius and pitch of the helix.
Answer:
\( r=\frac{mv\sin60^\circ}{eB}=\frac{1.67\times10^{-27}\times4\times10^5\times0.866}{1.6\times10^{-19}\times0.3}=1.2\times10^{-2}\ \text{m} \).
\( T=\frac{2\pi m}{eB}=2.19\times10^{-7}\ \text{s} \), so pitch \( =v\cos60^\circ\,T=2\times10^5\times2.19\times10^{-7}=4.4\times10^{-2}\ \text{m} \).

Illustration 6. An equilateral triangular coil of side 0.02 m carries 0.1 A and hangs between the poles of a magnet with a horizontal field of 0.05 T parallel to the plane of the coil. Find the torque.
Answer:
Area \( A=\frac{\sqrt3}{4}(0.02)^2=\sqrt3\times10^{-4}\ \text{m}^2 \) and moment \( m=IA=\sqrt3\times10^{-5}\ \text{A m}^2 \).
Field parallel to the plane means moment perpendicular to B: \( \tau=mB=\sqrt3\times10^{-5}\times0.05=5\sqrt3\times10^{-7}\ \text{N m} \).

Illustration 7. Show that the orbital magnetic moment of the electron in the ground state of hydrogen (Bohr model) is \( \frac{eh}{4\pi m} \) and find its direction.
Answer:
The moment of a circulating charge is \( \vec m=\frac{q}{2m}\vec L \), with q = -e, so \( \vec m=-\frac{e}{2m}\vec L \) (opposite to L).
Bohr's quantisation: \( L=\frac{nh}{2\pi} \), n = 1 for the ground state, so \( m=\frac{e}{2m}\cdot\frac{h}{2\pi}=\frac{eh}{4\pi m} \).