JEE Physics Moving Charges and Magnetism Notes Set 02

Revision Notes for JEE Physics: Moving Charges and Magnetic Field

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Force on Moving Charges and Currents

1. Lorentz force: \(\vec F=q\vec E+q(\vec v\times\vec B)\). The magnetic part does no work and never changes the speed. For a negative charge the force is opposite to \(\vec v\times\vec B\).
2. In a uniform field a charge moving perpendicular to \(\vec B\) goes on a circle: \(r=\dfrac{mv}{qB}\), \(T=\dfrac{2\pi m}{qB}\), frequency independent of speed and radius. At angle \(\theta\) to the field it follows a helix with radius \(r=\dfrac{mv\sin\theta}{qB}\) and pitch \(p=\dfrac{2\pi m\,v\cos\theta}{qB}\).
3. A particle must have \(r\ge\) width of the region (\(v\ge\dfrac{qBd}{m}\)) to cross a field region of width \(d\). With equal speeds in the same field, \(r\propto m/q\).
4. Straight conductor of length \(L\): \(\vec F=I\vec L\times\vec B\). Two parallel wires: \(\dfrac{F}{L}=\dfrac{\mu_0I_1I_2}{2\pi d}\), attraction for currents in the same direction and repulsion for opposite directions. In a uniform field the net force on a closed loop is zero, but its elements are pushed outwards or inwards.

Magnetic Field of Currents

SourceField
Long straight wire\(B=\dfrac{\mu_0I}{2\pi r}\)
Finite straight wire\(B=\dfrac{\mu_0I}{4\pi d}(\sin\alpha+\sin\beta)\); semi-infinite wire at the foot of the perpendicular: \(\dfrac{\mu_0I}{4\pi d}\)
Centre of circular coil of N turns\(B=\dfrac{\mu_0NI}{2R}\); arc of angle \(\phi\): \(\dfrac{\mu_0I\phi}{4\pi R}\)
Axis of circular coil\(B=\dfrac{\mu_0NIR^2}{2(R^2+x^2)^{3/2}}\approx\dfrac{\mu_0}{4\pi}\dfrac{2\mu}{x^3}\) for \(x\gg R\)
Solenoid / toroid\(B=\mu_0nI\); toroid \(B=\dfrac{\mu_0NI}{2\pi r}\)
Inside a solid wire (uniform current)\(B=\dfrac{\mu_0Ir}{2\pi R^2}\)
Rectangular loop, at its centre\(B=\dfrac{2\mu_0I}{\pi}\dfrac{\sqrt{a^2+b^2}}{ab}\)
Rotating charged ringequivalent current \(I=\dfrac{q\omega}{2\pi}\), \(B=\dfrac{\mu_0q\omega}{4\pi R}\)

Ampere's law: \(\oint\vec B\cdot d\vec l=\mu_0I_{enc}\). For a hollow cylindrical conductor, \(B=0\) inside the hole, it rises from zero at the inner surface to a maximum at the outer surface, and then falls as \(1/r\).

Magnetic Dipoles and the Earth's Field

1. Current loop: \(M=NIA\); torque \(\tau=MB\sin\theta\); potential energy \(U=-MB\cos\theta\). Work to turn a dipole from \(\theta_1\) to \(\theta_2\) is \(MB(\cos\theta_1-\cos\theta_2)\). Two perpendicular moments combine as \(\sqrt{M_1^2+M_2^2}\).
2. Bar magnet: axial field \(\dfrac{\mu_0}{4\pi}\dfrac{2M}{z^3}\) along \(\vec M\); equatorial field \(\dfrac{\mu_0}{4\pi}\dfrac{M}{z^3}\) opposite to \(\vec M\). In a non-uniform field a needle feels both a force and a torque.
3. Oscillation of a magnet: \(T=2\pi\sqrt{\dfrac{I}{MB}}\). Earth's field: \(B_H=B\cos\delta\), \(B_V=B\sin\delta\), \(\tan\delta=\dfrac{B_V}{B_H}\). The vertical component is zero at the magnetic equator. A dip needle vibrating in a horizontal plane feels only \(B_H\).
4. Tangent galvanometer: \(I=\dfrac{2RB_H}{\mu_0N}\tan\theta\).

Magnetic Materials

ClassSusceptibilityRelative permeabilityBehaviour
Diamagneticsmall, negative, independent of Tslightly below 1weakly repelled
Paramagneticsmall, positive, \(\chi\propto1/T\)slightly above 1weakly attracted
Ferromagneticlarge, positivemuch greater than 1strongly attracted; becomes paramagnetic above the Curie temperature

Hysteresis: retentivity is the magnetisation left at zero field and coercivity is the reverse field needed to remove it. Soft iron has a narrow loop (low retentivity, low coercivity), so it suits electromagnets and transformer cores. Steel has a wide loop with high retentivity and high coercivity, so it suits permanent magnets.

Worked Examples

1. Two long wires carrying \(10\ \text{A}\) and \(15\ \text{A}\) in the same direction, \(30\ \text{cm}\) apart: force on \(5\ \text{m}\) is \(2\times10^{-7}\times\dfrac{150\times5}{0.3}=5\times10^{-4}\ \text{N}\), attractive.
2. An electron of kinetic energy \(10\ \text{eV}\) has \(v=\sqrt{2K/m}\approx1.9\times10^6\ \text{m/s}\); in a field of \(10^{-4}\ \text{T}\) its orbit radius is \(\dfrac{mv}{eB}\approx0.1\ \text{m}\).
3. A semicircular loop of radius \(R\) joined to two semi-infinite straight wires gives at its centre \(B=\dfrac{\mu_0I}{4R}+2\dfrac{\mu_0I}{4\pi R}=\dfrac{\mu_0I}{4R}\left(1+\dfrac{2}{\pi}\right)\).

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