The horizontal component of the earth's magnetic field at a centain place is `3.0 xx 10^(-5)` T and the direction of the field is form the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1 A . What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west (b) south to north ?

Options

A. `3 xx 10^(-5) Nm^(-1), 0 Nm^(-1)`
B. `0 Nm^(-1) , 3xx 10^(-5) Nm^(-1)`
C. `0 Nm^(-1), 0 Nm^(-1)`
D. None of the above .

Answer

A

Explanation

Given `: B = 0.0 xx 10^(-5) T, l = 1 A`
Force per unit length is given by
`f = (F)/(l)`
where `F = I, B sin theta`
`:. f = I B sin theta`
(a) When the direction of current is from east to west , then `theta = 90^(@)`
Thus, force per unit length, f, is given by
`f = 1 A xx (3 xx 10^(-5) T) xx sin 90^(@)`
` = 3 xx 10^(-5) N m^(-1)`
The force `vec(f)` is directed in downward direction obtained using Fleming.s Left Hand Rule.
(b) When the direction of current is from South to North, then `theta = 0^(@)`
`:.` Thus ,
`f = 1 Axx (3 xx 10^(-5)T) xx sin 0^(@) = 0`
Hence, no force is acting on the conductor .