Magnetism and Magnetic Effects of Current
Magnetism and Magnetic Effects of Current for JEE Main & Advanced
Magnetic Field of Currents and Forces
Biot-Savart Law and Ampere's LawTopic 1
Biot-Savart Law: Magnetic field due to small current element $I\,d\vec{l}$ at a point distance $r$ away: $$d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$$ $\mu_0 = 4\pi \times 10^{-7}$ T·m/A (permeability of vacuum).
Magnetic Fields of Standard Currents:
| Configuration | Field magnitude |
|---|---|
| Straight wire (finite, angles $\theta_1, \theta_2$ from perpendicular at point) | $B = \dfrac{\mu_0 I}{4\pi a}(\sin\theta_1 + \sin\theta_2)$ |
| Infinite straight wire | $B = \dfrac{\mu_0 I}{2\pi a}$ |
| Circular loop (centre), radius $R$ | $B = \dfrac{\mu_0 I}{2R}$ |
| Loop on axis, distance $x$ | $B = \dfrac{\mu_0 IR^2}{2(R^2+x^2)^{3/2}}$ |
| Arc subtending angle $\theta$ at centre | $B = \dfrac{\mu_0 I\theta}{4\pi R}$ |
| Solenoid (long, inside) | $B = \mu_0 n I$ |
| Toroid (inside) | $B = \mu_0 n I = \mu_0 NI/(2\pi r)$ |
Ampere's Circuital Law: Line integral of $\vec{B}$ around any closed loop equals $\mu_0$ times enclosed current: $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$$
Useful for symmetric configurations like infinite wire, solenoid, toroid.
Right-hand rule: For straight wire, thumb along current → fingers curl in direction of $\vec{B}$. For a loop, fingers along current → thumb gives $\vec{B}$ at centre.
Find magnetic field at the centre of a circular coil of $50$ turns, radius $10$ cm, carrying current $2$ A.
Show solution
$$B = \frac{\mu_0 NI}{2R} = \frac{4\pi \times 10^{-7} \times 50 \times 2}{2 \times 0.1}$$ $$= \frac{4\pi \times 10^{-5}}{0.2} = 2\pi \times 10^{-4} \approx 6.28 \times 10^{-4} \text{ T}$$
Final Answer: $B \approx 6.28 \times 10^{-4}$ T.
A solenoid has $1000$ turns per metre and carries current $2$ A. Find magnetic field inside.
Show solution
$$B = \mu_0 n I = 4\pi \times 10^{-7} \times 1000 \times 2 = 8\pi \times 10^{-4} \approx 2.51 \times 10^{-3} \text{ T}$$
Final Answer: $B \approx 2.51 \times 10^{-3}$ T.
The unit of magnetic field is:
The magnetic field at the centre of a circular loop of radius $R$ carrying current $I$:
Ampere's law gives the magnetic field for:
The magnetic field inside an infinitely long solenoid:
The Biot-Savart law gives magnetic field due to:
Lorentz Force, Cyclotron and Force on WiresTopic 2
Lorentz Force on Moving Charge: $$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$$
Magnetic force is always perpendicular to velocity → does no work, changes only direction.
Motion of Charge in Magnetic Field:
- Velocity perpendicular to $\vec{B}$: circular motion of radius $r = mv/(qB)$, period $T = 2\pi m/(qB)$.
- Velocity at angle $\theta$ to $\vec{B}$: helical path with pitch $p = (v\cos\theta)T$.
- Time period independent of speed and radius.
Cyclotron: Accelerates charged particles using crossed electric and magnetic fields. Resonance condition: $$f_{\text{osc}} = \frac{qB}{2\pi m}$$ This frequency is independent of velocity and radius. Maximum KE: $K_{\max} = q^2B^2r^2/(2m)$.
Cyclotron limitation: At relativistic speeds, mass increases ($m_{\text{rel}}$), frequency changes — fixed-frequency cyclotron fails.
Force on Current-Carrying Wire in magnetic field: $$\vec{F} = I\vec{L}\times\vec{B}$$ Magnitude: $F = BIL\sin\theta$.
Force between Two Parallel Current-Carrying Wires: Wires separated by $d$ carrying $I_1, I_2$ in same direction attract; opposite directions repel. $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$ This relation defines the ampere: $1$ A is current that produces force of $2 \times 10^{-7}$ N/m between two parallel wires $1$ m apart.
Torque on Current Loop: A flat loop of area $A$ carrying current $I$ in field $B$: $$\vec{\tau} = \vec{M}\times\vec{B}$$ where $\vec{M} = NI\vec{A}$ is magnetic moment ($N$ = turns). Magnitude: $\tau = NIAB\sin\theta$.
Potential Energy: $U = -\vec{M}\cdot\vec{B} = -MB\cos\theta$.
An electron moves in a magnetic field $B = 1$ T at $v = 2 \times 10^6$ m/s perpendicular to $B$. Find radius of circular motion. ($m_e = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C)
Show solution
$$r = \frac{mv}{eB} = \frac{9.1 \times 10^{-31} \times 2 \times 10^6}{1.6 \times 10^{-19} \times 1}$$ $$= \frac{1.82 \times 10^{-24}}{1.6 \times 10^{-19}} \approx 1.14 \times 10^{-5} \text{ m}$$
Final Answer: $r \approx 1.14 \times 10^{-5}$ m $\approx 11.4\,\mu$m.
A wire of length $50$ cm carrying $2$ A is placed at $30°$ to a magnetic field of $0.5$ T. Find force.
Show solution
$$F = BIL\sin\theta = 0.5 \times 2 \times 0.5 \times 0.5 = 0.25 \text{ N}$$
Final Answer: $F = 0.25$ N.
The work done by magnetic force on a moving charge is:
A charged particle moves perpendicular to $\vec{B}$. Its trajectory is:
The period of revolution in cyclotron:
Two parallel wires carrying currents in same direction:
The torque on a current loop in a uniform field is:
Magnetic Properties of Matter
Earth's Magnetism and Magnetic MaterialsTopic 1
Earth's Magnetism: Earth acts like a bar magnet tilted at $\sim 11°$ to rotation axis. The south magnetic pole is in geographic north (since N-pole of compass points north).
Earth's Magnetic Field Components:
- Horizontal component $B_H = B_E\cos\delta$ (along magnetic meridian)
- Vertical component $B_V = B_E\sin\delta$
- Total intensity $B_E = \sqrt{B_H^2 + B_V^2}$
- Angle of dip $\delta$ = angle of $B_E$ with horizontal; $\tan\delta = B_V/B_H$
- Angle of declination: angle between magnetic and geographic meridian
At poles: $\delta = 90°$ (vertical). At equator: $\delta = 0°$ (horizontal).
Magnetic Quantities for Materials:
| Quantity | Definition | Symbol |
|---|---|---|
| Magnetic field intensity | applied | $\vec{H}$, units A/m |
| Magnetisation | magnetic moment per unit volume | $\vec{M}$ |
| Magnetic flux density | total field | $\vec{B}$ |
| Permeability | $\vec{B} = \mu\vec{H}$ | $\mu = \mu_0\mu_r$ |
| Susceptibility | $\vec{M} = \chi\vec{H}$ | $\chi = \mu_r - 1$ |
Types of Magnetic Materials:
| Type | $\chi$ | $\mu_r$ | Behaviour |
|---|---|---|---|
| Diamagnetic | Small, negative | < 1 | Weakly repelled (e.g., bismuth, water, copper) |
| Paramagnetic | Small, positive | > 1 | Weakly attracted (e.g., aluminium, oxygen) |
| Ferromagnetic | Large, positive | >> 1 | Strongly attracted; show hysteresis (e.g., iron, cobalt, nickel) |
Curie Law: For paramagnets, $\chi = C/T$ ($C$ = Curie constant). Above the Curie temperature, ferromagnets become paramagnetic.
Hysteresis Loop of ferromagnet: Area of $B$-$H$ loop represents energy dissipated per cycle. Soft iron: narrow loop, low retentivity, low coercivity — used in transformers and electromagnets. Steel: wide loop, high retentivity — used for permanent magnets.
At a place, $B_H = 0.3$ G and $B_V = 0.4$ G. Find the total Earth's magnetic field and angle of dip.
Show solution
$$B_E = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \text{ G}$$ $$\tan\delta = B_V/B_H = 0.4/0.3 = 4/3 \implies \delta \approx 53°$$
Final Answer: $B_E = 0.5$ G; $\delta \approx 53°$.
A magnetic material has susceptibility $\chi = 0.001$. Find $\mu_r$ and identify material type.
Show solution
$\mu_r = 1 + \chi = 1.001$ (slightly greater than 1). Material is paramagnetic.
Final Answer: $\mu_r = 1.001$; paramagnetic.
The angle of dip at the magnetic equator is:
Diamagnetic materials have:
Above Curie temperature, a ferromagnetic material becomes:
The hysteresis loop area represents:
Soft iron is used in electromagnets because:
Galvanometer, Moving Coil and Magnetic DipoleTopic 2
Moving Coil Galvanometer (MCG): A coil suspended in radial magnetic field. Torque on coil $\tau = NIAB$; balanced by torsion of suspension $\tau = C\phi$. So: $$\phi = \frac{NAB}{C}I$$ Deflection $\propto$ current.
Current sensitivity = $\phi/I = NAB/C$. Voltage sensitivity = $\phi/V = NAB/(CR)$.
Conversion to Ammeter: Connect low resistance shunt $S$ in parallel with galvanometer. $$S = \frac{I_g R_g}{I - I_g}$$ where $I_g$ = full-scale current of galvanometer, $R_g$ = galvanometer resistance, $I$ = desired full-scale current.
Conversion to Voltmeter: Connect high resistance $R$ in series with galvanometer. $$R = \frac{V}{I_g} - R_g$$
Magnetic Dipole: Closed current loop or bar magnet of moment $\vec{M} = NIA\hat{n}$ behaves like a dipole.
Bar magnet of length $2\ell$ and pole strength $m$: moment $M = m \cdot 2\ell$. Field on axis ($r \gg \ell$): $B_{\text{axial}} = (\mu_0/4\pi)(2M/r^3)$. On equatorial line: $B_{\text{eq}} = (\mu_0/4\pi)(M/r^3)$.
Comparison with electric dipole: Same formulas with $1/(4\pi\epsilon_0) \to \mu_0/(4\pi)$ and $p \to M$.
Oscillating bar magnet in uniform $\vec{B}$ (vibration magnetometer): $T = 2\pi\sqrt{I/(MB)}$.
A galvanometer of $R_g = 100$ Ω, $I_g = 1$ mA needs to read $10$ A as ammeter. Find shunt.
Show solution
$$S = \frac{I_g R_g}{I - I_g} = \frac{10^{-3} \times 100}{10 - 10^{-3}} \approx \frac{0.1}{10} = 0.01 \text{ Ω}$$
Final Answer: Shunt = $0.01$ Ω.
A bar magnet of moment $0.5$ A·m² is placed perpendicular to a uniform field $0.2$ T. Find torque.
Show solution
$$\tau = MB\sin\theta = 0.5 \times 0.2 \times \sin 90° = 0.1 \text{ N·m}$$
Final Answer: $\tau = 0.1$ N·m.
To convert a galvanometer to ammeter:
The voltage sensitivity of galvanometer increases if:
The magnetic moment of a current loop is:
Field on axis of a bar magnet at distance $r \gg \ell$:
The time period of oscillation of a magnet in a uniform field:
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