JEE Physics Moving Charges and Magnetism Notes Set 06

Magnetic Force on a Moving Charge

A charge q moving with velocity \( \vec v \) in a magnetic field \( \vec B \) feels \( \vec F=q\vec v\times\vec B \), so \( F=qvB\sin\theta \). The force is perpendicular to both v and B, hence it does no work and cannot change the speed or kinetic energy. With both fields present the Lorentz force is \( \vec F=q(\vec E+\vec v\times\vec B) \).
If v is perpendicular to B the particle moves on a circle of radius \( r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB} \), period \( T=\frac{2\pi m}{qB} \) (independent of speed) and angular frequency \( \omega=\frac{qB}{m} \). If v makes angle \( \theta \) with B, the path is a helix of radius \( \frac{mv\sin\theta}{qB} \) and pitch \( p=\frac{2\pi m}{qB}v\cos\theta \).
Crossed fields: a particle with \( v=\frac EB \) passes undeflected through perpendicular E and B (velocity selector). The cyclotron uses the independence of T from speed: the electric field in the gap is reversed every half revolution, so the radius grows as \( \sqrt n \) while the frequency stays fixed at \( \frac{qB}{2\pi m} \).

Biot-Savart Law and Fields of Simple Currents

A current element \( i\,d\vec l \) produces at a point with position vector \( \vec r \) from the element \( d\vec B=\frac{\mu_0}{4\pi}\frac{i\,d\vec l\times\vec r}{r^3} \), with \( \mu_0=4\pi\times10^{-7} \) T m/A. A moving point charge gives \( \vec B=\frac{\mu_0}{4\pi}\frac{q\vec v\times\vec r}{r^3} \), and \( \mu_0\varepsilon_0=\frac{1}{c^2} \).
Straight wire of finite length at perpendicular distance d, seen at angles \( \alpha \) and \( \beta \) from the perpendicular: \( B=\frac{\mu_0i}{4\pi d}(\sin\alpha+\sin\beta) \). For an infinite wire \( B=\frac{\mu_0i}{2\pi d} \); at the end of a semi-infinite wire, on the perpendicular through the end, \( B=\frac{\mu_0i}{4\pi d} \). Field lines are circles about the wire, with direction given by the right-hand grip rule.
Circular arc of radius R subtending angle \( \phi \) at its centre: \( B=\frac{\mu_0i\phi}{4\pi R} \), so a full loop gives \( \frac{\mu_0i}{2R} \) and N turns give \( \frac{\mu_0Ni}{2R} \). On the axis of a loop at distance x: \( B=\frac{\mu_0iR^2}{2(R^2+x^2)^{3/2}} \). Wires radial to the centre contribute nothing there, since \( d\vec l\parallel\vec r \).
Fields of different pieces add as vectors: for a conductor made of straight segments and arcs, find each contribution and its direction first, then add with signs.

Ampere's Law and Its Applications

\( \oint\vec B\cdot d\vec l=\mu_0 i_{enc} \), where the direction of the loop and the sign of the enclosed current follow the right-hand rule. B on the loop includes the contribution of all currents, inside or outside; only the enclosed current appears on the right.
Long straight wire of radius a with uniform current: inside \( B=\frac{\mu_0ir}{2\pi a^2} \), outside \( \frac{\mu_0i}{2\pi r} \). Hollow cylinder: zero inside, \( \frac{\mu_0i}{2\pi r} \) outside. For a non-uniform density J(r), integrate \( i_{enc}=\int J\,2\pi r\,dr \).
Long solenoid: \( B=\mu_0ni \) inside (uniform) and about zero outside; at the end of a long solenoid it is half this. Toroid with N turns: \( B=\frac{\mu_0Ni}{2\pi r} \) inside the core, zero outside and in the central hole. Infinite sheet with current per unit width K: \( B=\frac{\mu_0K}{2} \) on each side, directed parallel to the sheet and perpendicular to the current. A cylindrical cavity can be treated as the same cylinder with opposite current density added; inside the cavity the field is uniform.

Force on Current-Carrying Conductors

A straight segment in a uniform field: \( \vec F=i\vec L\times\vec B \), where \( \vec L \) is the vector from start to end of the segment. A curved wire in a uniform field feels the same force as the straight chord joining its ends. A closed loop in a uniform field has zero net force.
Two long parallel wires distance d apart: force per unit length \( \frac{\mu_0i_1i_2}{2\pi d} \), attractive for parallel currents and repulsive for antiparallel ones. A free wire above a fixed wire can be in equilibrium through this repulsion, and small vertical displacements x give simple harmonic motion with \( \omega^2=\frac gd \).
A flexible loop of radius R in a field perpendicular to its plane has tension \( T=iBR \) in the wire, from balancing the outward force on each element.

Magnetic Dipole Moment, Torque and Energy

A loop of area A with N turns and current i has moment \( \vec\mu=Ni\vec A \), directed by the right-hand rule. A rotating charge is equivalent to a current: a ring of charge q at angular speed \( \omega \) gives \( i=\frac{q\omega}{2\pi} \) and \( \mu=\frac{q\omega R^2}{2}=\frac{q}{2m}L \); a uniformly charged disc gives \( \frac{q\omega R^2}{4} \).
In a uniform field: \( \vec\tau=\vec\mu\times\vec B \), \( U=-\vec\mu\cdot\vec B \); equilibrium is stable when \( \vec\mu\parallel\vec B \) and unstable when antiparallel. Work to turn from \( \theta_1 \) to \( \theta_2 \) is \( \mu B(\cos\theta_1-\cos\theta_2) \). For small oscillations \( T=2\pi\sqrt{\frac{I}{\mu B}} \).
A moving-coil galvanometer uses this torque against a spring, with deflection proportional to current: \( NiAB=k\theta \). The sensitivity can be raised by more turns, a larger area, a stronger field or a weaker spring.

Worked Examples

Worked Example 1. A coil of 5 turns and radius 10 cm carries 1 A. Find the field at its centre.
Answer:
\( B=\frac{\mu_0NI}{2R}=\frac{4\pi\times10^{-7}\times5\times1}{2\times0.1} \)
\( B=3.14\times10^{-5} \) T.

Worked Example 2. An electron moves at \( 10^6 \) m/s perpendicular to a field of 0.01 T. Find the radius and period of its circular path.
Answer:
\( r=\frac{mv}{eB}=\frac{9.1\times10^{-31}\times10^6}{1.6\times10^{-19}\times0.01}=5.7\times10^{-4} \) m.
\( T=\frac{2\pi m}{eB}=\frac{2\pi\times9.1\times10^{-31}}{1.6\times10^{-19}\times0.01}=3.6\times10^{-9} \) s, independent of the speed.

Worked Example 3. Two long parallel wires 5 cm apart each carry 10 A in the same direction. Find the force per metre between them.
Answer:
Field of one at the other: \( B=\frac{\mu_0i}{2\pi d}=\frac{2\times10^{-7}\times10}{0.05}=4\times10^{-5} \) T.
Force per metre \( =iB=10\times4\times10^{-5}=4\times10^{-4} \) N/m, attractive.

Worked Example 4. A long solenoid has 1000 turns per metre and carries 2 A. Find B inside, and the torque on a 50-turn coil of area 4 cm\( ^2 \) carrying 0.5 A with its plane parallel to the axis.
Answer:
\( B=\mu_0ni=4\pi\times10^{-7}\times1000\times2=2.5\times10^{-3} \) T.
The plane is parallel to B, so \( \tau=NiAB=50\times0.5\times4\times10^{-4}\times2.5\times10^{-3}=2.5\times10^{-5} \) N m.

Worked Example 5. A proton and an alpha particle are accelerated through the same potential difference and enter a uniform magnetic field perpendicular to it. Find the ratio of the radii.
Answer:
\( r=\frac{\sqrt{2mqV}}{qB}\propto\sqrt{\frac mq} \).
\( \frac{r_p}{r_\alpha}=\sqrt{\frac{m_p/e}{4m_p/2e}}=\frac{1}{\sqrt2} \).