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 $$1.5$$×$$10$$−$$2$$ T. A current of $$10.0$$ ampere flow 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

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  $$1.5$$×$$10$$−$$2$$ T. A current of $$10.0$$ ampere flow 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

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 by τ$$=NIBA$$ $$sin$$θWhere  $$I=current$$ in the coil,  θ$$=angle$$ which the normal to the plane of the coil makes with the lines of induction B. Here,  $$N=1$$ ; $$B=1.5$$×$$10$$−$$2T$$ $$A=0.05$$×$$0.08=40$$×$$10-4$$ $$m2I=10.0amp$$,θ$$=90$$°$$=$$π$$/2$$∴τ$$=1.5$$×$$10-2(10.0)$$×$$(1)40$$×$$10-4sin($$π$$/2)=6$$×$$10-4Nm$$.

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