Magnetic Effects of Current & Magnetism

The Biot-Savart law is the magnetic analog of Coulomb's law.
It gives the magnetic field dB produced by a small current element Idl at a point P at distance r: dB = ($\frac{\mu_{0}}{4}$ $\pi$)(Idl sin $\theta$ / $r^{2}$), where $\mu_{0}$ = 4 $\pi$ x $10^{-7}$ T m/A is the permeability of free space; [$\mu_{0}$] = [M L $T^{-2} \; A^{-2}$]; [B] = [M $T^{-2} \; A^{-1}$], SI unit: tesla (T). The direction of dB is given by the right-hand rule: curl fingers from dl toward r-hat, and the thumb points along dB.

For an infinitely long straight conductor: B = $\mu_{0}$ I/(2 $\pi$ d), where d is the perpendicular distance; [B] = [M $T^{-2} \; A^{-1}$], SI unit: T.
For a finite wire, B = ($\mu_{0}$ I/4 $\pi$ d)(sin $\alpha$ + sin $\beta$). The field forms concentric circles around the wire; direction follows the right-hand thumb rule (thumb along current, curled fingers give B direction).

At the center of a circular loop of N turns and radius R carrying current I: B = $\mu_{0}$ NI/(2R).
On the axis at distance x: B = $\mu_{0} \; NIR^{2}$ / [2($R^{2}$ + $x^{2}$)^($\frac{3}{2}$)].
For a solenoid: B = $\mu_{0}$ nI (inside, uniform), where n = N/L is the number of turns per unit length; outside the solenoid, B is approximately zero.

Ampere's circuital law: The line integral of B along a closed loop equals $\mu_{0}$ times the enclosed current: $\oint B .<br/>dl$ = $\mu_{0} \; I_{enc}$. This is most useful with high symmetry (infinite wire, solenoid, toroid).
For a toroid: B = $\mu_{0}$ NI/(2 $\pi$ r) inside the toroid, zero outside.

The Lorentz force on a charge q moving with velocity v in electric field E and magnetic field B is F = q(E + v x B).
The magnetic component F = qvB sin $\theta$ is always perpendicular to v, so it does NO work — it changes direction, not speed.
A charge moving perpendicular to B follows a circular path: radius r = mv/(qB); [r] = [L], SI unit: m.
The time period T = 2 $\pi$ m/(qB) is independent of velocity and radius; [T] = [T], SI unit: s.
The frequency f = qB/(2 $\pi$ m).
If v has a component along B, the motion is helical with pitch = $v_{parallel}$ x T.

Force on a current-carrying conductor: F = BIl sin $\theta$; [F] = [M L $T^{-2}$], SI unit: N.
Force between parallel conductors: F/l = $\mu_{0} \; I_{1} \; I_{2}$/(2 $\pi$ d); same-direction currents attract, opposite currents repel. This defines the ampere: 1 A produces a force of 2 x $10^{-7}$ N/m between two infinite parallel wires 1 m apart.

Torque on a current loop in a magnetic field: $\tau$ = NIAB sin $\theta$ = MB sin $\theta$, where M = NIA is the magnetic moment; [M] = [A $L^{2}$], SI unit: A $m^{2}$; [$\tau$] = [M $L^{2} \; T^{-2}$], SI unit: N m.

The moving coil galvanometer uses torque on a current loop: deflection $\theta$ = (NAB/k)I, where k is the torsional constant of the spring.
Current sensitivity = $\theta$/I = NAB/k.
To convert to an ammeter, connect a low-resistance shunt S = $I_{g}$ G/(I - $I_{g}$) in parallel.
To convert to a voltmeter, connect a high resistance R = V/$I_{g}$ - G in series.

Magnetic materials: Diamagnetic ($\chi$ < 0, $\mu_{r}$ < 1; examples: Cu, Bi, $H_{2O}$ — weakly repelled by field), Paramagnetic ($\chi$ > 0 small, $\mu_{r}$ slightly > 1; examples: Al, $O_{2}$ — weakly attracted), Ferromagnetic ($\chi$ >> 0, $\mu_{r}$ >> 1; examples: Fe, Co, Ni — strongly attracted).
Curie's law: $\chi$ = C/T for paramagnets. Above the Curie temperature, ferromagnets become paramagnetic. The hysteresis loop shows retentivity (residual magnetism) and coercivity (field needed to demagnetize): soft ferromagnets (low coercivity, electromagnets) vs hard ferromagnets (high coercivity, permanent magnets).

The key testable concept is the magnetic field due to various current configurations (Biot-Savart and Ampere's law) combined with the force on charges and conductors in magnetic fields, which together dominate NEET questions from this chapter.

Solved Numericals

N₁. A circular coil of 100 turns, radius 10 cm carries a current of 2 A. Find the magnetic field at the center and at a point 10 cm along the axis.

Given: N = 100, R = 10 cm = 0.10 m, I = 2 A, $\mu_{0}$ = 4 $\pi$ x $10^{-7}$ T m/A.

At center (x = 0): $B_{center}$ = $\mu_{0}$ NI/(2R) = (4 $\pi$ x $10^{-7}$ T m/A x 100 x 2 A) / (2 x 0.10 m) = (4 $\pi$ x $10^{-7}$ x 200) / 0.20 T = (800 $\pi$ x $10^{-7}$) / 0.20 T = 4000 $\pi$ x $10^{-7}$ T = 4 $\pi$ x $10^{-4}$ T = 1.257 x $10^{-3}$ T = 1.257 mT.

At axial point x = 10 cm = 0.10 m = R: $B_{axis}$ = $\mu_{0} \; NIR^{2}$ / [2($R^{2}$ + $x^{2}$)^($\frac{3}{2}$)] = (4 $\pi$ x $10^{-7}$ x 100 x 2 x (0.10)²) / [2 x (0.01 + 0.01)^($\frac{3}{2}$)] = (4 $\pi$ x $10^{-7}$ x 200 x 0.01) / [2 x (0.02)^($\frac{3}{2}$)] Numerator = 4 $\pi$ x $10^{-7}$ x 2 = 8 $\pi$ x $10^{-7}$. (0.02)^($\frac{3}{2}$) = (0.02)¹ x (0.02)^($\frac{1}{2}$) = 0.02 x 0.1414 = 2.828 x $10^{-3}$. Denominator = 2 x 2.828 x $10^{-3}$ = 5.657 x $10^{-3}$. $B_{axis}$ = 8 $\pi$ x $10^{-7}$ / 5.657 x $10^{-3}$ = 8 $\pi$ x $10^{-4}$ / 5.657 = 25.13 x $10^{-4}$ / 5.657 = 4.44 x $10^{-4}$ T = 0.444 mT.

The field at the axial point (x = R) is $B_{center}$/(2 $\sqrt{2}$) = 1.$\frac{257}{2}$.828 = 0.444 mT. Verified.

N₂. A proton (m = 1.67 x $10^{-27}$ kg, q = 1.6 x $10^{-19}$ C) enters a uniform magnetic field B = 0.1 T perpendicular to its velocity of $10^{6}$ m/s. Find the radius, time period, and frequency.

Radius: r = mv/(qB) = (1.67 x $10^{-27}$ kg x $10^{6}$ m/s) / (1.6 x $10^{-19}$ C x 0.1 T) = 1.67 x $10^{-21}$ / 1.6 x $10^{-20}$ m = 0.104 m = 10.4 cm.

Time period: T = 2 $\pi$ m/(qB) = 2 $\pi$ x 1.67 x $10^{-27}$ / (1.6 x $10^{-19}$ x 0.1) = 2 $\pi$ x 1.67 x $10^{-27}$ / 1.6 x $10^{-20}$ = 2 $\pi$ x 1.044 x $10^{-7}$ s = 6.56 x $10^{-7}$ s = 0.656 us.

Frequency: f = 1/T = 1/(6.56 x $10^{-7}$) = 1.524 x $10^{6}$ Hz = 1.524 MHz.

Note: T and f are independent of velocity. A faster proton traces a larger circle in the same time.

N₃. A galvanometer with resistance G = 50 ohm gives full-scale deflection for $I_{g}$ = 1 mA. Find (a) shunt to convert to ammeter of range 0-5 A, and (b) series resistance for voltmeter of range 0-10 V.

(a) Ammeter (shunt in parallel): S = $I_{g}$ G / (I - $I_{g}$) = (1 x $10^{-3}$ A x 50 ohm) / (5 - 0.001) A = 0.050 / 4.999 ohm = 0.01 ohm (approximately).

Effective resistance of ammeter = GS/(G + S) = 50 x 0.01 / 50.01 = 0.01 ohm (very low, as required).

(b) Voltmeter (series resistance): R = V/$I_{g}$ - G = 10 V / (1 x $10^{-3}$ A) - 50 ohm = 10000 - 50 = 9950 ohm.

Effective resistance of voltmeter = G + R = 50 + 9950 = 10000 ohm = 10 k-ohm (very high, as required).