Magnetic Field due to Electric Current

The magnetic field due to an electric current is calculated using the Biot-Savart law, which gives the field of an infinitesimal current element as $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{\ell }\times \hat{r}}{r^2}$. Integrating this over the geometry gives the field of any current configuration. For a JEE Main and NEET student, the four essential results are: field of a long straight wire ($B=\frac{{\mu }_{0}I}{2\pi r}$), of a finite wire ($B=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$), of a circular arc at its centre ($B=\frac{{\mu }_{0}I\varphi }{4\pi R}$), and of a circular loop on its axis. These four cover almost every JEE and NEET question on this topic.

1Biot-Savart Law

Consider a wire carrying steady current $I$. Divide it into infinitesimal length vectors $d\vec{\ell }$ pointing along the direction of current. The magnetic field $d\vec{B}$ produced at a point $P$ at position $\vec{r}$ from the element is given by Biot-Savart's law:

Here ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$ is the permeability of free space, and $\frac{{\mu }_0}{4\pi }=10^{-7}\text{T⋅m/A}$. The magnitude is $dB=\frac{{\mu }_0}{4\pi }\frac{Id\ell \sin \theta }{r^2}$, where $\theta$ is the angle between $d\vec{\ell }$ and $\vec{r}$.

Direction (right-hand rule)

The direction of $d\vec{B}$ is that of $d\vec{\ell }\times \hat{r}$. Practically: point the right-hand thumb along the current $I$; the curl of the fingers gives the direction of $\vec{B}$ around the wire.

2Magnetic Field due to a Straight Current-Carrying Wire

2.1Finite wire (general result)

For a straight wire of finite length, let $P$ be a point at perpendicular distance $d$ from the wire. Let ${\theta }_1$ and ${\theta }_2$ be the angles subtended at $P$ by the two ends, measured from the perpendicular $OP$ dropped from $P$ onto the wire. Then

Sign convention: ${\theta }_1$ and ${\theta }_2$ are taken positive when measured on opposite sides of the foot of the perpendicular; if both ends lie on the same side, one of them is negative.

2.2Special cases

• Infinite wire: ${\theta }_1={\theta }_2=90°$, so $B=\frac{{\mu }_{0}I}{2\pi d}$. Field lines are concentric circles around the wire.
• Semi-infinite wire (one end at foot of perpendicular): ${\theta }_1=0°$, ${\theta }_2=90°$, giving $B=\frac{{\mu }_{0}I}{4\pi d}$.
• Point on the extension of the wire: $d\vec{\ell }\parallel \hat{r}$ so $d\vec{\ell }\times \hat{r}=0$, hence $B=0$.

3Field at the Centre of a Circular Loop

For a circular loop of radius $R$ carrying current $I$, every element $d\vec{\ell }$ is perpendicular to $\vec{r}$ (which points from the element to the centre), so $\sin \theta =1$ and each element contributes $dB=\frac{{\mu }_0}{4\pi }\frac{Id\ell }{R^2}$ in the same direction (perpendicular to the loop). Integrating over $d\ell =2\pi R$:

3.1Field due to a circular arc

An arc of radius $R$ subtending angle $\varphi$ (in radians) at the centre carries the same reasoning but integrates over $d\ell =Rd\varphi$ from $0$ to $\varphi$:

• Semicircle ($\varphi =\pi$): $B=\frac{{\mu }_{0}I}{4R}$
• Quarter circle ($\varphi =\pi /2$): $B=\frac{{\mu }_{0}I}{8R}$
• Full loop ($\varphi =2\pi$): $B=\frac{{\mu }_{0}I}{2R}$ (recovers loop result)

4Field on the Axis of a Circular Loop

For a point $P$ on the axis of a circular loop at distance $x$ from the centre, each element $Id\vec{\ell }$ contributes $dB=\frac{{\mu }_0}{4\pi }\frac{Id\ell }{r^2}$ where $r=\sqrt{R^2+x^2}$. By symmetry, only the component along the axis survives (the sine of the angle between $\vec{r}$ and the axis gives $R/r$):

4.1Special cases

• Centre ($x=0$): $B=\frac{{\mu }_{0}NI}{2R}$ (recovers loop-centre result).
• Far from loop ($x\gg R$): $B\approx \frac{{\mu }_{0}NI{R}^2}{2x^3}=\frac{{\mu }_0}{4\pi }\frac{2M}{x^3}$, where $M=NIA=NI\pi R^2$ is the magnetic dipole moment of the loop. This matches the axial field of a magnetic dipole - the loop behaves as a tiny bar magnet.

5Direction Conventions Summary

Common Mistakes to Avoid