Magnetism formula sheet — Biot-Savart, Ampere, force on wire, torque on loop

Question

A circular loop of radius 10 cm carries a current of 5 A. Find the magnetic field at the centre of the loop and the magnetic moment of the loop.

(JEE Main / NEET / CBSE 12 — Moving Charges and Magnetism)

Magnetism Problem Classification

flowchart TD
    A["Magnetism Problem"] --> B{What to find?}
    B -->|"B field from a wire/loop"| C{Geometry?}
    B -->|"Force on a current-carrying wire"| D["F = BIL sin theta"]
    B -->|"Torque on a loop"| E["tau = NIAB sin theta = M x B"]
    B -->|"Force between two wires"| F["F/L = mu_0 I1 I2 / (2 pi d)"]
    C -->|Straight wire| G["Biot-Savart or Ampere"]
    C -->|Circular loop| H["B = mu_0 I / (2R) at centre"]
    C -->|Solenoid| I["B = mu_0 n I"]
    G --> G1["Infinite wire: B = mu_0 I / (2 pi r)"]

Solution — Step by Step

The formula for the magnetic field at the centre of a circular loop carrying current $I$ :

B = \frac{\mu_0 I}{2R}

Given: $I = 5$ A, $R = 10$ cm = 0.1 m, $\mu_0 = 4\pi \times 10^{-7}$ T $\cdot$ m/A

B = \frac{4\pi \times 10^{-7} \times 5}{2 \times 0.1} = \frac{20\pi \times 10^{-7}}{0.2}

\boxed{B = 10\pi \times 10^{-6} \text{ T} = 31.4 \times 10^{-6} \text{ T} \approx 31.4 \text{ } \mu\text{T}}

Magnetic moment: $M = NIA$

For a single loop ( $N = 1$ ):

M = 1 \times 5 \times \pi(0.1)^2 = 5 \times 0.01\pi

\boxed{M = 0.05\pi \text{ A}\cdot\text{m}^2 \approx 0.157 \text{ A}\cdot\text{m}^2}

The magnetic moment direction follows the right-hand rule: curl fingers in the direction of current, thumb points in the direction of $\vec{M}$ .

Situation	Formula
Field at centre of loop	$B = \mu_0 I/(2R)$
Field on axis of loop (distance $x$ )	$B = \mu_0 IR^2/[2(R^2 + x^2)^{3/2}]$
Infinite straight wire (distance $r$ )	$B = \mu_0 I/(2\pi r)$
Solenoid (inside)	$B = \mu_0 nI$ ( $n$ = turns/length)
Toroid	$B = \mu_0 NI/(2\pi r)$ ( $N$ = total turns)
Force on wire	$\vec{F} = I\vec{L} \times \vec{B}$
Torque on loop	$\vec{\tau} = \vec{M} \times \vec{B}$

Why This Works

The Biot-Savart law gives the magnetic field from any current element: $d\vec{B} = \frac{\mu_0}{4\pi}\frac{Id\vec{l} \times \hat{r}}{r^2}$ . Integrating this over a circular loop gives the centre-field formula. Ampere’s law ( $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$ ) provides a shortcut when there is symmetry — similar to how Gauss’s law shortcuts electrostatics.

Alternative Method — Force Between Parallel Wires

Two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ :

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

Currents in the same direction attract; opposite directions repel. This is the basis for the SI definition of the ampere.

For JEE/NEET, the most asked magnetism numericals involve: (1) field at the centre of a loop, (2) force between parallel wires, and (3) torque on a current loop in a uniform field. Master these three and you cover 60% of the problems from this chapter.

Common Mistake

Students frequently confuse the formula for the field of a straight wire ( $\mu_0 I/2\pi r$ ) with the field at the centre of a loop ( $\mu_0 I/2R$ ). Notice: the straight wire formula has $\pi$ in the denominator, the loop formula does not. Mixing them up gives an answer off by a factor of $\pi$ — which is easy to catch if you check units and orders of magnitude.