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What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 30^@ with the direction of a uniform magnetic field of 0.15 T ? |
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Answer» Solution :Here `I = 8 A, theta = 30^@ and B = 0.15 T` `:.` Force per unit LENGTH of the wire `= F/l = B I SIN theta = 0.15 xx 8 xx sin 30^@ = 0.15 xx 8 xx 1/2 = 0.6 N m^(-1)`. |
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