A straight conducting rod of length `30 cm` and having a resistance of `0.2` ohm is allowed to slide over two parallel thick metallic rails with uniform velocity of `0.2 m//s` as shown in the figure. The rails are situated in a horizontal plane if the horizontal component of earth's magnetic field is `0.3 xx 10^(-4)T` and a steady current of `3 muA` is induced through the rod. The angle of dip will be:

Options

A. `tan^(-1)((3)/(4))`
B. `tan^(-1)((1)/(sqrt(3)))`
C. `tan^(-1) (sqrt(3))`
D. `tan^(-1)((1)/(3))`

Answer

D

Explanation

Only vertical lines of earth's magnetism will cut when the rod is allowed to slide over the rails. Let the vertical component of earht's magnetic field is `B_(v)`. Induced
current `i = (B_(v)l upsilon-0)/(R ) = (B_(v)l upsilon)/(R )`
`B_(V) = 10^(-5)T` Given `B_(H) = 0.3 xx 10^(-4) T`
The angle of dip `theta` is given by `theta = tan^(-1)((B_(v))/(B_(H)))`