JEE Physics Moving Charges and Magnetism Notes Set 01

Magnetic Field of a Current: Biot-Savart Law

A current element \( I\,d\vec l \) sets up a field at a point with position vector \( \vec r \) (from the element) given by \[ d\vec B=\frac{\mu_0}{4\pi}\,\frac{I\,d\vec l\times\vec r}{r^3} \] where \( \mu_0/4\pi=10^{-7} \) T m/A. The direction follows the right-hand rule: point the thumb along the current and the fingers curl along the field. The unit of B is the tesla (1 T = 10⁴ gauss).

Standard Results

Infinite straight wire at perpendicular distance r: \( B=\frac{\mu_0I}{2\pi r} \).
Finite straight wire seen at angles \( \alpha,\beta \) from the perpendicular: \( B=\frac{\mu_0I}{4\pi r}(\sin\alpha+\sin\beta) \).
Centre of a circular coil of N turns: \( B=\frac{\mu_0NI}{2R} \); for an arc subtending angle \( \theta \): \( B=\frac{\mu_0I\theta}{4\pi R} \).
Axis of a coil at distance x from the centre: \( B=\frac{\mu_0NIR^2}{2(R^2+x^2)^{3/2}} \), which behaves as \( 1/x^3 \) for \( x\gg R \).
Magnetic moment of a loop: \( M=NIA \), so on the axis far away \( B=\frac{\mu_0}{4\pi}\frac{2M}{x^3} \).

Ampere's Circuital Law and Its Uses

The line integral of B around any closed path equals \( \mu_0 \) times the net current threading the path: \( \oint\vec B\cdot d\vec l=\mu_0I_{enc} \). Use it when the current distribution is symmetric.
Long solenoid (inside): \( B=\mu_0nI \), uniform; outside it is almost zero. At the end the field is half this value.
Toroid of N turns and mean radius r: \( B=\frac{\mu_0NI}{2\pi r} \) inside the core, and zero outside.
Thick cylindrical wire of radius a: \( B=\frac{\mu_0Ir}{2\pi a^2} \) for r less than a, and \( \frac{\mu_0I}{2\pi r} \) for r greater than a.

Force on a Moving Charge

The Lorentz force is \( \vec F=q(\vec E+\vec v\times\vec B) \). The magnetic part is perpendicular to the velocity, so it does no work and never changes the speed.
Velocity perpendicular to B: circular motion with \( r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB} \), time period \( T=\frac{2\pi m}{qB} \) (independent of speed).
Velocity at angle θ to B: a helix with radius \( \frac{mv\sin\theta}{qB} \) and pitch \( \frac{2\pi mv\cos\theta}{qB} \).
Velocity selector: undeflected when \( qE=qvB \), i.e. \( v=E/B \).
Cyclotron: positive ions are accelerated across a gap by an alternating field of frequency \( f=\frac{qB}{2\pi m} \); the maximum kinetic energy is \( \frac{q^2B^2R^2}{2m} \).

Force on Conductors and Between Wires

A straight wire of length l in a uniform field feels \( \vec F=I\,\vec l\times\vec B \). A bent wire behaves like a straight wire joining its ends, and a closed loop in a uniform field has zero net force.
Two long parallel wires distance d apart attract if the currents are parallel, and repel if antiparallel, with force per unit length \( \frac{\mu_0I_1I_2}{2\pi d} \). This defines the ampere: 1 A in each of two long wires 1 m apart gives \( 2\times10^{-7} \) N per metre.

Loops, Torque and Galvanometers

A loop of N turns and area A with current I has moment \( \vec M=NI\vec A \). In a field B it feels a torque \( \vec\tau=\vec M\times\vec B \), of size \( NIAB\sin\theta \) with \( \theta \) the angle between the normal and B. The potential energy is \( U=-\vec M\cdot\vec B \).
A moving-coil galvanometer balances the magnetic torque against a spring torque: \( NIAB=k\phi \), so the deflection is proportional to the current. Current sensitivity is \( \phi/I=NAB/k \); voltage sensitivity is the same divided by the coil resistance.
To convert it into an ammeter connect a small shunt \( S=\frac{I_gG}{I-I_g} \) in parallel; to make a voltmeter connect a large series resistance \( R=\frac VI_g-G \).
For an electron orbiting with angular momentum L the moment is \( \mu=\frac{eL}{2m} \); with \( L=n\hbar \) the smallest unit is the Bohr magneton \( 9.27\times10^{-24} \) A m².

Solved Examples

Illustration 1. A long straight wire carries 5 A. Find the magnetic field at 10 cm from it.
Answer:
\( B=\frac{\mu_0I}{2\pi r}=\frac{2\times10^{-7}\times5}{0.1}=10^{-5} \) T.
B = 10 microtesla.

Illustration 2. A circular coil of 100 turns and radius 5 cm carries 0.5 A. Find the field at its centre.
Answer:
\( B=\frac{\mu_0NI}{2R}=\frac{4\pi\times10^{-7}\times100\times0.5}{2\times0.05} \)
B = 6.28 × 10⁻⁴ T.

Illustration 3. A proton is accelerated through 1000 V and enters a field of 0.1 T perpendicular to its motion. Find the radius of its path (m = 1.67 × 10⁻²⁷ kg).
Answer:
\( v=\sqrt{2eV/m}=\sqrt{\frac{2\times1.6\times10^{-19}\times1000}{1.67\times10^{-27}}}\approx4.38\times10^{5} \) m/s.
\( r=\frac{mv}{eB}=\frac{1.67\times10^{-27}\times4.38\times10^{5}}{1.6\times10^{-19}\times0.1}\approx0.0457 \) m.
r is about 4.6 cm.

Illustration 4. Two long parallel wires 20 cm apart carry 10 A and 20 A in the same direction. Find the force per metre.
Answer:
\( \frac Fl=\frac{2\times10^{-7}\times10\times20}{0.2}=2\times10^{-4} \) N/m.
2 × 10⁻⁴ N per metre, attractive.

Illustration 5. A rectangular coil of 50 turns, 4 cm × 5 cm, carries 2 A and lies with its plane parallel to a field of 0.2 T. Find the torque.
Answer:
\( \tau=NIAB\sin90^\circ=50\times2\times(0.04\times0.05)\times0.2 \)
Torque = 0.04 N m.