Physics Magnetic Effects of Current and Magnetism

Magnetic Effects of Current and Magnetism Formula Sheet

All key magnetism formulas for JEE - Biot-Savart, Ampere's law, Lorentz force, torque, dipoles, galvanometers, and magnetic materials. Quick revision for JEE Main & Advanced.

2 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know formula from this chapter in one scannable place. Use it for last-minute revision: each result below is drawn directly from the chapter topics on Biot-Savart law, Ampere’s law, Lorentz force, force on conductors, galvanometers, magnetic dipoles, and magnetic materials.

Master constant

Memorize $\mu_0 = 4\pi \times 10^{-7}$ T·m/A and the shortcut $\dfrac{\mu_0}{4\pi} = 10^{-7}$ T·m/A. The second form saves real time in almost every numerical.

Biot-Savart Law and Magnetic Field of Currents

Field due to a current element:

$$\boxed{d\vec{B} = \frac{\mu_0}{4\pi}\cdot\frac{I\,d\vec{l}\times\vec{r}}{r^3}} \qquad dB = \frac{\mu_0}{4\pi}\cdot\frac{I\,dl\sin\theta}{r^2}$$
ConfigurationMagnetic fieldKey point
Infinite straight wire$B = \dfrac{\mu_0 I}{2\pi r}$Most common building block
Finite wire (angles $\alpha,\beta$)$B = \dfrac{\mu_0 I}{4\pi r}(\sin\alpha + \sin\beta)$Measured from the ends
Semi-infinite wire$B = \dfrac{\mu_0 I}{4\pi r}$Half of the infinite wire
Circular loop (center)$B = \dfrac{\mu_0 N I}{2R}$$\perp$ to plane of loop
Circular loop (on axis)$B = \dfrac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}$The $3/2$ power
Loop, far on axis ($x \gg R$)$B \approx \dfrac{\mu_0 I R^2}{2x^3} = \dfrac{\mu_0 M}{2\pi x^3}$$M = I\pi R^2$
Arc subtending $\theta$ at center$B = \dfrac{\mu_0 I \theta}{4\pi R}$$\theta$ in radians
Square loop, side $a$ (center)$B = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi a}$Sum of 4 sides

Arc special cases (from $B = \dfrac{\mu_0 I\theta}{4\pi R}$):

$$\text{Semicircle: } B = \frac{\mu_0 I}{4R} \qquad \text{Quarter: } B = \frac{\mu_0 I}{8R} \qquad \text{Full circle: } B = \frac{\mu_0 I}{2R}$$
Direction shortcut

Right-hand thumb rule: point the thumb along the current, the curled fingers give the circular field direction. The field around a straight wire forms loops, it never points radially.

Ampere’s Circuital Law

$$\boxed{\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}}$$

Maxwell-corrected form (with displacement current):

$$\oint \vec{B}\cdot d\vec{l} = \mu_0\left(I_{\text{enc}} + \epsilon_0\frac{d\Phi_E}{dt}\right)$$
ConfigurationRegionMagnetic field
Infinite wireOutside$B = \dfrac{\mu_0 I}{2\pi r}$
Solid cylinder (radius $R$)Inside, $r$B = \dfrac{\mu_0 I r}{2\pi R^2}$
Solid cylinderOutside, $r>R$$B = \dfrac{\mu_0 I}{2\pi r}$
SolenoidInside$B = \mu_0 n I$
SolenoidOutside (ideal)$B = 0$
ToroidInside$B = \dfrac{\mu_0 N I}{2\pi r}$
ToroidOutside$B = 0$
Infinite current sheetEither side$B = \dfrac{\mu_0 K}{2}$

Hollow cylinder (inner $a$, outer $b$), current within the conductor ($a $$B = \frac{\mu_0 I (r^2 - a^2)}{2\pi r (b^2 - a^2)} \qquad (rb:\ B=\tfrac{\mu_0 I}{2\pi r})$$

When to use which law

Use Ampere’s law only when there is cylindrical, planar, or translational symmetry (infinite wire, solenoid, toroid, cylinder, current sheet). For finite or bent wires, loops, and arcs, fall back to Biot-Savart.

Lorentz Force and Motion of Charged Particles

$$\boxed{\vec{F} = q(\vec{v}\times\vec{B})} \qquad F = qvB\sin\theta$$

Complete Lorentz force (electric + magnetic):

$$\boxed{\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})}$$
Magnetic force does NO work

The magnetic force is always perpendicular to $\vec{v}$, so $W = \vec{F}\cdot\vec{v} = 0$. The magnetic field changes a particle’s direction, never its speed or kinetic energy.

QuantityFormulaNote
Radius (circular, $v\perp B$)$r = \dfrac{mv}{qB}$Larger mass and speed widen the circle
Time period$T = \dfrac{2\pi m}{qB}$Independent of $v$ and $r$
Cyclotron frequency$f = \dfrac{qB}{2\pi m}$Independent of $v$
Angular frequency$\omega = \dfrac{qB}{m}$Simplest form
Radius (helical)$r = \dfrac{mv\sin\theta}{qB}$Uses $v_\perp$ only
Pitch (helical)$p = \dfrac{2\pi m v\cos\theta}{qB}$Uses $v_\parallel$

Motion summary by angle between $\vec{v}$ and $\vec{B}$: $\theta = 0^\circ$ gives a straight line ($F=0$); $\theta = 90^\circ$ gives a circle; any other angle gives a helix.

Applications (Crossed and Cyclotron Fields)

DeviceKey formulaNote
Velocity selector$v = \dfrac{E}{B}$Only this speed passes undeflected
Cyclotron AC frequency$f = \dfrac{qB}{2\pi m}$Independent of radius
Cyclotron max KE$KE_{\max} = \dfrac{q^2 B^2 R^2}{2m}$At dee radius $R$
Mass spectrometer$m = \dfrac{qB^2 r}{E}$After velocity selection $v=E/B$
Magnetic focusing$d = \dfrac{\pi m v}{qB}$Distance between focuses
$E \parallel B$accelerating helixHelix with increasing pitch
$E \perp B$ (from rest)drift $v_{\text{drift}} = \dfrac{E}{B}$Cycloidal motion

Particles accelerated through the same potential $V$ then bent in field $B$:

$$r = \frac{1}{B}\sqrt{\frac{2mV}{q}} \quad\Rightarrow\quad \frac{r_1}{r_2} = \sqrt{\frac{m_1}{m_2}} \ \ (\text{same } q,V,B)$$

Force and Torque on Current-Carrying Conductors

$$\boxed{d\vec{F} = I\,d\vec{l}\times\vec{B}} \qquad \boxed{\vec{F} = I\vec{L}\times\vec{B}} \qquad F = BIL\sin\theta$$
QuantityFormulaNote
Force on straight wire$F = BIL\sin\theta$Max ($BIL$) when $I\perp B$
Force between parallel wires$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$Same direction $\to$ attract
Torque on coil$\tau = NIAB\sin\theta$Max when plane $\parallel B$
Magnetic moment$M = NIA$$\vec{\tau} = \vec{M}\times\vec{B}$
Potential energy$U = -\vec{M}\cdot\vec{B} = -MB\cos\theta$Min at $\theta = 0^\circ$
Work to rotate$W = MB(\cos\theta_1 - \cos\theta_2)$$\theta_1\to\theta_2$
Effective length theorem

For any shape of wire in a uniform field, the force equals that on a straight wire joining the endpoints: $\vec{F} = I\vec{L}_{\text{eff}}\times\vec{B}$. So a semicircle uses $L_{\text{eff}} = 2R$ (diameter), and a closed loop has $L_{\text{eff}} = 0$, hence zero net force (but possibly nonzero torque).

Common sign and direction traps

Same-direction currents attract, opposite-direction currents repel. Torque is zero at equilibrium ($\theta=0$), maximum at $\theta=90^\circ$. The potential energy keeps its negative sign: $U = -MB\cos\theta$.

Definition of the ampere: two infinite parallel wires 1 m apart each carrying 1 A experience a force of $2\times 10^{-7}$ N/m.

Moving Coil Galvanometer

QuantityFormulaNote
Deflection$\theta = \dfrac{NBA}{k}\,I$Linear; $\propto I$ (radial field)
Current sensitivity$S_I = \dfrac{\theta}{I} = \dfrac{NBA}{k}$Higher is more sensitive
Voltage sensitivity$S_V = \dfrac{\theta}{V} = \dfrac{NBA}{kG}$Depends on $G$ too
Figure of merit$k_g = \dfrac{1}{S_I}$Lower is better

Conversion to ammeter (shunt $S$ in parallel, $n = I/I_g$):

$$\boxed{S = \frac{I_g G}{I - I_g} = \frac{G}{n-1}} \qquad R_A = \frac{GS}{G+S} = \frac{G}{n}$$

Conversion to voltmeter (resistance $R$ in series, $m = V/(I_g G)$):

$$\boxed{R = \frac{V}{I_g} - G = (m-1)G} \qquad R_V = G + R = mG = \frac{V}{I_g}$$
PropertyAmmeterVoltmeter
AddShunt in parallel (low R)Resistance in series (high R)
Ideal resistance$R_A \to 0$$R_V \to \infty$
Quality$\dfrac{R_V}{V} = \dfrac{1}{I_g}$ ($\Omega$/V)
Half-deflection method

To find the galvanometer resistance, halve the deflection by adding a shunt; at half-deflection $G = S$, because equal currents flow through equal parallel resistances.

Magnetic Dipole and Bar Magnet

$$\boxed{\vec{M} = NI\vec{A}} \qquad M = m\times 2l \ (\text{bar magnet})$$
QuantityFormulaNote
Torque$\tau = MB\sin\theta$Max at $\theta = 90^\circ$
Potential energy$U = -MB\cos\theta$Min (stable) at $\theta = 0^\circ$
Work to rotate$W = MB(\cos\theta_1 - \cos\theta_2)$$\theta_1\to\theta_2$
Energy gap (stable$\to$unstable)$\Delta U = 2MB$$0^\circ \to 180^\circ$
Oscillation period$T = 2\pi\sqrt{\dfrac{I}{MB}}$Like a torsion pendulum
Angular frequency$\omega = \sqrt{\dfrac{MB}{I}}$$I$ = moment of inertia
Axial field ($r\gg$ length)$B = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$Along axis
Equatorial field$B = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$Opposite to $\vec{M}$
Field ratio$\dfrac{B_{\text{axial}}}{B_{\text{equatorial}}} = 2$At same distance

General point at angle $\theta$ to the axis:

$$B_r = \frac{\mu_0}{4\pi}\frac{2M\cos\theta}{r^3}, \quad B_\theta = \frac{\mu_0}{4\pi}\frac{M\sin\theta}{r^3}, \quad B = \frac{\mu_0}{4\pi}\frac{M}{r^3}\sqrt{1+3\cos^2\theta}, \quad \tan\alpha = \frac{\tan\theta}{2}$$

Force between two coaxial dipoles falls as $r^{-4}$: $F = \dfrac{6\mu_0 M_1 M_2}{4\pi r^4}$ (coaxial), $F = \dfrac{3\mu_0 M_1 M_2}{4\pi r^4}$ (side by side).

Bohr magneton: $\mu_B = \dfrac{e\hbar}{2m_e} = 9.27\times 10^{-24}$ A·m².

Dipole pitfalls

Axial field is twice the equatorial field (not the other way around). The energy sign is negative: $U = -MB\cos\theta$. Cutting a magnet always yields two magnets - no isolated monopoles - and cutting in half either way gives moment $M/2$.

Earth’s Magnetic Field

QuantityRelationNote
Angle of dip$\tan\theta = \dfrac{B_V}{B_H}$$0^\circ$ at equator, $90^\circ$ at poles
Horizontal component$B_H = B\cos\theta$
Vertical component$B_V = B\sin\theta$

Earth’s moment $M_E \approx 8\times 10^{22}$ A·m²; field near equator $B_E \approx 3\times 10^{-5}$ T (30 μT).

Magnetic Materials

QuantityFormulaNote
Magnetic intensity$\vec{H} = \dfrac{\vec{B}}{\mu_0} - \vec{M}$Unit A/m
Solenoid intensity$H = nI$Depends only on current
Magnetization$M = \chi H$Moment per unit volume
Susceptibility$\chi = \mu_r - 1$Dimensionless
Relative permeability$\mu_r = 1 + \chi$$\mu_r = \mu/\mu_0$
B-H relation$B = \mu_0\mu_r H$In a medium
B-M relation$B = \mu_0(H + M) = \mu_0(1+\chi)H$Alternative form
Curie law (para)$\chi = \dfrac{C}{T}$$\chi \propto 1/T$
Curie-Weiss (ferro)$\chi = \dfrac{C}{T - T_C}$Valid above $T_C$
Hysteresis loss$E = \oint H\,dB$Loop area = energy/cycle/volume

Classification of Materials

PropertyDiamagneticParamagneticFerromagnetic
Susceptibility $\chi$Negative ($\sim -10^{-5}$)Small positive ($10^{-5}$ to $10^{-3}$)Large positive ($10^2$ to $10^5$)
Permeability $\mu_r$$<1$$>1$ (slightly)$\gg 1$ (1000 to 100000)
Behavior in fieldWeakly repelledWeakly attractedStrongly attracted
Temperature effectIndependent$\chi \propto 1/T$$\chi \propto 1/(T-T_C)$
ExamplesBi, Cu, Ag, Au, H₂OAl, Na, O₂, PtFe, Co, Ni

Curie temperatures: Iron 1043 K, Cobalt 1394 K, Nickel 631 K. Above $T_C$ a ferromagnet becomes paramagnetic.

Hysteresis Terms

  • Retentivity (remanence): the residual $B$ when $H = 0$ (field is removed).
  • Coercivity: the reverse magnetizing intensity $H$ needed to bring the magnetization back to zero.
  • Saturation: all domains aligned, $M = M_s$; further increase in $H$ no longer increases $M$.
PropertySoft magneticHard magnetic
Hysteresis loopNarrow (small area)Wide (large area)
Retentivity / CoercivityLowHigh
Energy lossLowHigh
UseTransformer cores, electromagnetsPermanent magnets, data storage
ExamplesSoft iron, permalloySteel, alnico, NdFeB
Material misconceptions

Diamagnetic $\chi$ is negative. The permeability relation is $\mu_r = 1 + \chi$ (plus sign). Paramagnetic $\chi$ decreases with temperature. Domains exist always in a ferromagnet; the external field merely aligns them.

Chapter Map

graph TD
    A[Magnetism] --> B[Field of currents]
    A --> C[Forces and torque]
    A --> D[Dipoles]
    A --> E[Materials]
    B --> B1["Biot-Savart: dB = (mu0/4pi) I dl x r / r^3"]
    B --> B2["Ampere: closed integral B.dl = mu0 I_enc"]
    C --> C1["Lorentz: F = q(v x B)"]
    C --> C2["Conductor: F = I L x B"]
    C --> C3["Galvanometer: theta = NBA I / k"]
    D --> D1["M = NIA, tau = M x B, U = -M.B"]
    D --> D2["Axial = 2 x Equatorial field"]
    E --> E1["chi = mu_r - 1, B = mu0(H+M)"]
    E --> E2["Curie & hysteresis"]

Last-Minute Reminders

High-yield recall list

  • $\dfrac{\mu_0}{4\pi} = 10^{-7}$ T·m/A and $\mu_0 = 4\pi\times10^{-7}$ T·m/A.
  • Cyclotron period $T = \dfrac{2\pi m}{qB}$ and frequency are independent of speed and radius.
  • Magnetic force does no work; speed and KE stay constant.
  • Same-direction currents attract; opposite repel.
  • Torque on a loop is zero at equilibrium, maximum when the plane is parallel to $B$.
  • Axial dipole field is twice the equatorial field at the same distance.
  • Ammeter: low-resistance shunt in parallel; voltmeter: high-resistance in series.
  • $\mu_r = 1 + \chi$; diamagnetic $\chi < 0$, paramagnetic small positive, ferromagnetic $\chi \gg 0$.

Further Reading