5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30 \degreewith the direction of a uniform magnetic field of 0.15 T?

Answers (1)
S Sayak

The magnetic force on an infinitesimal current-carrying conductor in a magnetic field is given by \vec{dF}=I\vec{dl}\times \vec{B} where the direction of vector dl is in the direction of flow of current.

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 force per unit length is equal to

|F| = 0.15\times8\timessin30o         (I=8A, B=0.15 T, \theta=30o)

=0.6 Nm-1

 

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