A conductor (shown in the figure) carrying constant current `I` is kept in the `x-y` plane in a uniform magnetic field ` vec(B)`. If ` vec(F)` is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are)

Options

A. If `vec B " is along " hat z, F propto (L + R)`
B. If `vec B " is along " hat x , F = 0 `
C. If `vec B " is along " hat y , F propto (L + R)`
D. If `vec B " is " along " , hat z , F = 0 `

Answer

a, b, c

Explanation

Force on a current carrying wire placed inside a uniform magnetic field is given by: `vec F = I (vec l xx vec B)`
`vec l` in the above equation, just represents a straight length vector drawn between two ends of conductor that is not straight.
Magnitude of `vec l` in the given situation is equal to `(2L + 2R)`.
The direction of the electric current is along the positive direction of X-axis in the wire.
The magnetic force acts on the wire only when magnetic field is applied along any direction except in the direction of current (here, X - axis).
`F = I l B sin theta`
` = I B (2R + 2L) sin theta`
If the direction of magnetic field is along positive X-axis, the direction of length and magnetic field are in the same direction so net force is zero because `sin theta = 0 `.
Thus, option (b) is correct
If the magnetic field is along Y- or Z-axis, then the angle between `vec l " and " vec B " is " 90^(@)`. Thus,
`F = IB l sin 90^(@)`
`= IB (2R + 2L)`
` F propto (R + L)`
Therefore, options (a) and (c ) are also correct.