Applications of Biot Savart's Law —
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Applications of Biot Savart's Law —

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Applications of Biot Savart's Law —

Magnetic field due to a straight current carrying wire

The magnetic field $B$ at a point $P$ due to a straight wire of finite length carrying current $I$ at a perpendicular distance $r$ is

$B=\frac{\mu_0 I}{4 \pi r}[\sin \alpha+\sin \beta]$

Special cases :

  • If the wire is of infinite length and the point $P$ lies near the centre of the straight wire, then

$\alpha=\beta=90^{\circ}, \quad \therefore \quad B=\frac{\mu_0 2 I}{4 \pi r}=\frac{\mu_0 I}{2 \pi r}$

  • If the wire is of infinite length and the point $P$ lies near one end, then

  $\alpha=\frac{\pi}{2}, \beta=0^{\circ}, \therefore B=\frac{\mu_0 I}{4 \pi r}$

The direction of magnetic field due to a straight current carrying wire is given by right hand thumb rule. According to this rule, if you grasp the wire in your right hand with your extended thumb pointing in the direction of the current. Your fingers curling around the wire give the direction of the magnetic field lines.

Magnetic field at the centre of a current carrying circular coil

The magnetic field at the centre of a circular coil of radius $a$ carrying current $I$ is

$B=\frac{\mu_0}{4 \pi} \frac{2 \pi I}{a}=\frac{\mu_0 I}{2 a}$

If the circular coil consists of $N$ turns, then

$B=\frac{\mu_0}{4 \pi} \frac{2 \pi N I}{a}=\frac{\mu_0 N I}{2 a}$

The direction of magnetic field at the centre of a circular coil carrying current is given by right hand rule.

According to which, the direction of magnetic field at the centre of the circular coil is perpendicular to the plane of coil downwards for the clockwise current, and perpendicular to the plane of coil outwards for the anticlockwise current.

The magnetic field at the common centre $\mathrm{O}$ of two concentric coils of radii $a$ and $b$ having turns $N_1$ and $N_2$ in which the same current $I$ is flowing

  • when current flows in the same direction

$B=\frac{\mu_0 I}{2}\left[\frac{N_1}{a}+\frac{N_2}{b}\right]$

  • when current flows in the opposite direction

$B=\frac{\mu_0 I}{2}\left[\frac{N_1}{a}-\frac{N_2}{b}\right]$

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