Shows a metallic wire of resistance `0.20 Omega` sliding on a horizontal, U shaped metallic rail The separation between the parallel arms is 20 cm. An electric current of `2.0 mu A` passes through the wire when it is slid at a rate of `20cm s ^(-1)`. If the horizontal component of the earth's magnetic field is `3.0 X 10^(-5)` T, calculate the dip at the place.

Explanation

Here `l=20cm=20xx10^(-2)m`
`V=20cm//s`
`=20xx10^(-2)m//s`
`B_H=3xx10^(-5)T,`
`i=2 mu A=2xx10^(-6) A,`
`R=0.2 Omega.`
`i=(B_0 lV)/(r)`
` implies B_v=(iR)/(lV)`
`=(2xx10^(-5)xx2xx10^(-1))/(20xx10^(-2)xx20xx10^(-2))`
`=(2xx2xx10^(-7))/(2xx2xx10^(-2))`
`=1xx10^(-5) Tesla.`
`tan delta=(B_v)/(B_H)=(1xx10^(-5))/(3xx10^(-5))=1/3`
`implies delta(dip)=tan^(-1)(1/3).`