Question

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

Answer 1

Question 30905

The magnetic field is directed along the axis of solenoid. General formula for Amperes law is $d\vec{F} = I[\vec{d} l,\vec{B} ]$ , but for given wire $\vec{F} = I[\vec{l},\vec{B} ]$ since wire is straight ( $\vec{l}$ is directed along the wire, and $|\vec{l}| = l$ ). Opening the cross product gives sine of the angle, which is equal to 1 because angle between $\vec{l}$ and $\vec{B}$ is 90 degrees.

Hence, the magnitude of the force is $|F_{A}| = B I l \sin 90 = 0.27 T \cdot 10 A \cdot \frac{3}{100} m = 0.081 N$ .