Question

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

Answer 1

Given: The length of the wire is 3 cm , the current carried by the wire is 10 A , the uniform magnetic field is 0.27 T , and the angle between the magnetic field and the wire is 90°.

Magnetic force on wire is given as,

F=BIlsinθ

where, the magnetic field is B, current carried in the wire is I, length of the wire is l , and the angle between the wire and the magnetic field is θ.

By substituting the given values in the above equation, we get,

F=0.27×10×3 cm×( 1 m 100 cm )×sin90° =8.1× 10 −2  N

Therefore, the magnetic force on the wire is 8.1× 10 −2  N.