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A coil area 100 cm^(2) having 500 tums carries a current of 1 mA . It is suspended in a uniform magnetic field of induction 10^(-3) Wb//m^(2). Its plane makes an angle of 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=100cm^(2)=100xx10^(-4)m^(2)` COUPLE acting on the coil is given by `tau=BiANsinphi,` Where `phi` is angle made by NORMAL to the PLANE of coil with B. `phi =90 - 60 = 30°` `thereforeC=10^(-3)xx10^(-3)xx100xx10^(-4)xx500xxsin30` `=250xx10^(-8)Nm` |
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