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A coil of area 100 cm^2 having 500 tuns carries a current of 1 mAl. It is suspended in a uniform magnetic field of induction10^(-3)Wb//m^2 . Its plane makes an angleof 60^@ with the lines of induction. Find the torque acting on the coil. |
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Answer» Solution :Given i = 1 `mA = 10^(-3)` A , N= 500`B = 10^(-3) Wb//m^2` `theta = 60^@ , tau = ? A = 100 cm^2 = 100 xx 10^(-4) m^2` Couple acting on the COIL is given by `tau = BiA N SIN phi` where `phi` is angle made by normal to the plane of coil with B. `phi = 90 - 60 = 30^@` `therefore C = 10^(-3) xx 10^(-3) xx 100 xx 10^(-4) xx 500 xx sin 30^@` `= 250 xx 10^(-8)` Nm |
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