A uniform rod of length L and mass M is hinged at its upper point and is at rest at that moment in vertical plane. current i flows in it. A uniform magnetic field B exists perpendicular to the rod in the hirizontal direction. Find the force exerted by the hinge on the rod just after the release.

Please answer!!!

force due to magnetic field = ILBsin(theta)

=iLBsin 90

=iLB ( direction will depend on the direction of current but it will be in plane of paper either towards left or right)

other force acting on the rod = T ( due to hinge )

and also due to gravity = Mg

for balancing

vector(T)=vector(Mg)+vector(iLB)

Hence your answer can be calculated by plugging in the given values.

Thanks & Regards

Saurabh Singh,

askIITians Faculty

B.Tech.

IIT Kanpur