Find the magnetic field due to a semi-infinite length wire at point $P$ as shown in the figure.

  • A
    $B_p = \frac{{\mu _0}I}{{4\pi d}}[\sin \theta + 1]$
  • B
    $B_p = \frac{{\mu _0}d}{{4\pi I}}[\sin \theta ]$
  • C
    $B_p = \frac{{\mu _0}I}{{4\pi d}}[\sin \theta + \sin \theta ]$
  • D
    $B_p = \frac{{\mu _0}I}{{4\pi d}}[\sin 90^\circ + \sin 90^\circ]$

Explore More

Similar Questions

Do magnetic forces obey Newton's third law? Verify for two current elements $\overrightarrow{dl_1} = dl(\hat{i})$ located at the origin and $\overrightarrow{dl_2} = dl(\hat{j})$ located at $(0, R, 0)$. Both carry current $I$.

An electron revolves around a nucleus with rotational frequency $f$ in a circular orbit. Due to this,the magnetic induction produced at the nucleus position is $B$. The radius of the circular orbit is directly proportional to:

In a wire of radius $1 \ mm$,a steady current of $2 \ A$ uniformly distributed across the cross-section of the wire is flowing. Then the magnetic field at a point $0.25 \ mm$ from the centre of the wire is (in $\mu T$)

$A$ current $i$ is flowing in a straight conductor of length $L.$ The magnetic induction at a point on its axis at a distance $\frac{L}{4}$ from its centre will be

The magnetic field due to a straight conductor of uniform cross-section of radius $a$ and carrying a steady current is represented by

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