A metal ring of radius $$0.5m$$ with its plane normal to a uniform magnetic field B of induction $$0.2T$$ carries a current $$100A$$ (clockwise). The tension in newton developed in the ring is

Question

Infinity Learn · Accepted Answer

Given B perpendicular to the plane of the ring,  from the figure direction of magnetic force is in positive $$Y-direction$$ for upper semicircular part.                      Magnetic force on upper semicircular part ,   F→$$=idl$$→ X B→$$=$$ (i)×(2r)×$$Bj^=100$$ $$2$$×$$0.5$$ X $$210j^=20j^$$ So, balancing force on $$semi-circular$$ ring we get,  Tension in two arms $$=$$ Magnetic force on upper half ⇒$$2T$$ $$=$$ $$20$$ ⇒T $$=$$ $$10$$ N

A metal ring of radius 0.5m with its plane normal to a uniform magnetic field B of induction 0.2T carries a current 100A (clockwise). The tension in newton developed in the ring is

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