A coil of area 100 cm^2 having 500 turns carries a current of 1 m A. It is suspended in a uniform magnetic field of induction 10^-3wb/m^2. Its plane making an angle of 60° with the lines of induction. the torque acting on the coil is
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Solution
T=BAIN cos theeta B =10^-3 A=100cm2=100*10^-4 m2 I =1mA=1*10^-3A N=500 theeta =60
so T =( 10^-3)*(10^-2)*(10^-3)*500*cos60 =250*10^-8 Nm =2.5*10^-6 Nm