A conducting rod of length l and mass m is moving down a smooth inclined plane of inclination `theta` with constant velocity V as shown in figure. A current I is flowing in the conductor in a direction perpendicular to paper inwards. Let a vertically upward magnetic field `vecB` be existent in space. Then, what is the magnitude of magnetic field `vecB`?

Options

A. `(mg)/(il) sin (theta)`
B. `(mh)/(il) "tan" (theta)`
C. `(mg cos (theta))/(il)`
D. `(mg)/(il sin (theta))`

Answer

B

Explanation

The magnetic force due to the conducting rod of length I and mass m is,
`F_m = i (vecl xx vecB),` the angle between l and B is `90^@`
` F_m = ilB`
The horizontal component of `F_m` is
`F^1 = F_m Cos theta`
The force due to gravity is `F= mg `
The horizontal component of force due to gravity is
`F^H = mg sin theta`
Rod will move downward with constant velocity if,
`F^1 = F^H`
`implies F_m Cos theta = mgSin theta`
`implies il B Cos theta = mg Sin theta`
`implies B = (mg)/(il)(Sin theta)/(Cos theta) = (mg)/(il) "tan" theta`