$A$ straight wire of length $2 \ m$ carries a current of $10 \ A$. If this wire is placed in a uniform magnetic field of $0.15 \ T$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be

  • A
    $1.5 \ N$
  • B
    $3 \ N$
  • C
    $3 \sqrt{2} \ N$
  • D
    $\frac{3}{\sqrt{2}} \ N$

Explore More

Similar Questions

An isosceles triangular current-carrying loop is placed in a uniform magnetic field $\overrightarrow{B_{o}}$ directed perpendicular to the plane of the loop,as shown in the figure. The net magnetic force on the loop is:

$A$ current of $5 \ A$ is flowing in a wire of length $1.5 \ m$. $A$ force of $7.5 \ N$ acts on it when it is placed in a uniform magnetic field of $2 \ T$. The angle between the magnetic field and the direction of the current is......$^o$

$A$ wire carrying a steady current $I$ is kept in the $x$-$y$ plane along the curve $y=A \sin \left(\frac{2 \pi}{\lambda} x\right)$. $A$ magnetic field $B$ exists in the $z$-direction. The magnitude of the magnetic force on the portion of the wire between $x=0$ and $x=\lambda$ is

$A$ conducting wire $PQ$ carries a current $10 \ A$ as shown in the figure. It is placed in a uniform magnetic field $5 \ T$ which is acting normally outwards from the paper. The net force experienced by the wire is:

Two parallel conductors carrying unequal currents in the same direction . . . . . .

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