1.

A wheel with 10 metallic spokes each 0.5 m long rotated with a speed of 120 "rev"/"min" in a plane normal to the horizontal component of earth's magnetic field B_h at a place. If B_h = 0.4G at the place, what is the induced emf between the axle and the rim of the wheel ? (1 G = 10^(-4) T)

Answer»

0V
0.628 mV
0.628 `muV`
`62.8 muV`

Solution :Induced EMF PRODUCED between the axle and the rim of the WHEEL is
`epsilon=(B omegaR^2)/2`…(i)
(Here B=0.4 , `G=0.4 xx10^(-4)` T)
`omega=120 "rev"/"MIN"=(240pi)/60"rad"/s`
`omega=4pi` rad/s,R=0.5 m, N=10
`THEREFORE epsilon=(0.4xx10^(-4)xx4xx3.14xx(0.5)^2)/2` [ `because` From eqn. (i)]
`=(0.4xx4xx3.14xx0.25xx10^(-4))/2`
`=0.628xx10^(-4)`
`=62.8xx10^(-6)`
`=62.8 muV`


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