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 `vecB` is along `hatz`, `Fprop(L+R)`
B. If `vecB` is along `hatx`, `F=0`
C. If `vecB` is along `haty`, `Fprop(L+R)`
D. If `vecB` is along `hatz`, `F=0`

Answer

A, B, C

Explanation

While finding the magnetic force on a current carrying wire in magnetic field, the vector length of the wire is taken. Hence, the vector length of the wire is
`vecl=(L+R+R+L)hati=2(L+R)hati`
Force on conductor of length l, carrying current I when placed in a magnetic field `vecB=(Bhatk)` is
`vecF=I[veclxxvecB]=I[2(L+R)hatixxBhatk]`
`=2IB(L+R)(-hatj)`
`:. Fprop(L+R)`
When `vecB` is along x-axis i.e., `vecB=Bhati`, then
`vecF=I[2(L+R)hatixxBhati]=0` (`:'hatixxhati=0`)
When `vecB` is along y-axis i.e., `vecB=Bhatj`, then
`vecF=I[2(L+R)hatixxBhatj]=2I(L+R)Bhatk`
So, `Fprop(L+R)`