A horizontal rod of mass 10 gm and length 10 cm is placed on a smooth plane inclined at an angle of 60 degree with the horizontal, with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction B is applied vertically downwards. If the current through the rod is 1.73 ampere, then the value of B for which the rod remains stationary on the inclined plane is

  • Option 1)

    1.73 Tesla
     

  • Option 2)

    1 / 1.73Tesla

  • Option 3)

    1 Tesla

     

  • Option 4)

    None of the above

 

Answers (1)

As we learnt in

Total magnetic force -

\underset{F}{\rightarrow}=i ( \underset{L}{\rightarrow}\times \underset{B}{\rightarrow})

- wherein

 

 

 

For equilibrium

mg sin \Theta = i l B cos \Theta

 \Rightarrow 10^{-2} \times 10 \times\frac{\sqrt{3}}{2}=1.73 \times 0.1 \times B \times\frac{1}{2}

\Rightarrow    B = 1 T


Option 1)

1.73 Tesla
 

incorrect 

Option 2)

1 / 1.73Tesla

incorrect 

Option 3)

1 Tesla

 

correct 

Option 4)

None of the above

incorrect 

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