Question

A $3$ cm wire carrying current of $10A$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27$ $T.$ What is the magnetic force on the wire ?

• A. $8.1\times {10}^{-2}N$
• B. $9.1\times {10}^{-2}N$
• C. $0N$
• D. $25.43\times {10}^{-2}N$

Answer 1

The correct option is A $8.1\times {10}^{-2}N$

$Given:$ Length of the wire, $l=3cm=0.03m$

Current flowing in the wire, $I=10A$

Magnetic field, $B=0.27T$

$Solution:$ Angle between the current and magnetic field, $\Omega ,=$

${90}^{\circ }$

Magnetic force exerted on the wire is given as:

$F=BIl\sin \Omega$

$=0.27\times 10\times 0.03\sin {90}^{\circ }$

$=8.1\times {10}^{-2}N$

Hence, the magnetic force on the wire is $8.1\times {10}^{-2}N$.

The direction of the force can be obtained from

Fleming's left hand rule.

$So,the$ $correct$ $option:A$