6. A 3.0 cm wire carrying a current of 10 A 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?

Answers (1)
S Sayak

For a straight wire of length l in a uniform magnetic field, the Force equals to

\\\vec{F}=\int_{0}^{l}I\vec{dl}\times \vec{B}\\ |\vec{F}|=BIlsin\theta

In the given case the magnitude of the force is equal to

|F| = 0.27\times10\times0.03\timessin90o         (I=10A, B=0.27 T, \theta=90o)

=0.081 N

The direction of this force depends on the orientation of the coil and the current-carrying wire and can be known using the Flemings Left-hand rule.

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