Question

A $3.0cm$ wire carrying a current of $10A$ is a placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27T$. What is magnetic force on the wire?

Answer 1

Given,
$i=10A,l=3cm=0.03m,$

$B=0.27T\text{and}\theta ={90}^{\circ }$

Magnetic force on a current carrying conductor is given by:
$\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})=ilB\sin \theta$

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

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

Direction by force is finding out by Fleming left hand rule.


Final Answer: $8.1\times {10}^{-2}N$; direction of force given by Fleming's left-hand rule