The plane of a rectangular loop of wire with sides $$0.05$$ m and $$0.08$$ m is parallel to a uniform magnetic field of induction $$1.5$$ x $$10-2$$ T. A current of $$10.0$$ ampere flows through the loop. If the side of length $$0.08$$ m is normal and the side of length $$0.05$$ m is parallel to the lines of induction, then the torque acting on the loop is

Question

Infinity Learn · Accepted Answer

Torque τ acting on a current carrying coil of area A placed in a magnetic field of induction B is given byτ $$=$$ NIBA $$sin$$ θwhere .I $$=$$ current in the coil , θ $$=$$ angle which the normal to theplane of the coil makes with the lines of induction B.Here,$$N=1$$, $$B=1.5$$×$$10-2$$ T $$A=0.05$$×$$0.08=40$$×$$10-4$$ $$m2$$ $$I=10.0amp$$,θ$$=90$$°$$=$$π$$/2$$ τ$$=iBANsin$$θ τ$$=1.5$$×$$10-2(10.0)$$×(1)40$$×$$10-4sin($$π$$/2) $$=6$$×$$10-4Nm$$

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