A metallic rod of mass per unit length `0.5 kg m^(-1)` is lying horizontally on a straght inclined plane which makes an angle of `30^(@)` with the horizontal. The rod is not allowed to slide down by flowing a current throguh it when a magnetic field of induction `0.25 T` is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

Options

A. 14.76 A
B. 5.98 A
C. 7.14 A
D. 11.32 A

Answer

D

Explanation

d) Key concep, Firstly, make a free body diagram of the system and indicate the magnitude and direction of all the forces acting on the body, Then, choose any two mutually perpendicualr axes say X and Y in the plane of forces in case of coplanar forces. As the systm is in equilibrium,
`SigmaF_(x)=0` or `mgsintheta=Fcostheta`............(i)
Where, F is the magnitude of force experienced by the rod when placed in a magnetic field and current I is flowing through it.
But the force expereienced by the given rod in a unifom magnetic field is
F=ILB
`therefore` Eq. (i) becomes
`mgsintheta=ILBcostheta`
`rArr I = (mgsintheta)/(LBcostheta)=(mg)/(LB)tantheta`
`I= (m/L)(gtantheta)/B`..........(ii)
According to the question.

Here, `m/L= 0.5kgm^(-1), g=9.8 ms^(-2), theta=30^(@)`, B=0.25 T
Substituting the given values in Eq. (ii), we get
`I=(0.5 xx 9.8)/(0.25)tan30^(@)` = (0.5 xx 9.8)/(0.25) xx 1/sqrt(3) == 11.32` A