Discuss special cases of the Biot-Savart law.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $(1)$ If $\theta = 0^{\circ}$,then $\sin 0^{\circ} = 0$,which implies $dB = 0$. This means the magnetic field is zero at points located on the axis of the current element.
$(2)$ If $\theta = 90^{\circ}$,then $\sin 90^{\circ} = 1$,which implies $dB$ is maximum. This means the magnetic field due to a current element is maximum in a plane passing through the element and perpendicular to its axis.

Explore More

Similar Questions

Surface charge density on a ring of radius $a$ and width $d$ is $\sigma$ as shown in the figure. It rotates with frequency $f$ about its own axis. Assume that the charge is only on the outer surface. The magnetic field induction at the centre is (Assume that $d \ll a$):

$A$ current $i$ flows in a circular arc of wire of radius $R$,which subtends an angle of $3\pi / 2$ radians at its centre. The magnetic induction at the centre is

$A$ circular coil of wire of radius $r$ has $n$ turns and carries a current $I$. The magnetic induction $B$ at a point on the axis of the coil at a distance $\sqrt{3} r$ from its centre is

Two points $A$ and $B$ on the axis of a circular current loop are at distances of $4 \ cm$ and $3 \sqrt{3} \ cm$ from the centre of the loop. If the ratio of the induced magnetic fields at points $A$ and $B$ is $216: 125$, the radius of the loop is (in $cm$)

$A$ charge $q$ moving in a circle of radius $r$ metre makes $n$ revolutions per second. The magnetic field at the centre of the circle is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo