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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
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