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  • On a smooth plane inclined at 30°C with the horizontal, a thin current-carrying metallic rod is placed parallel to the horizontal ground. The plane is in a uniform magnetic field of 0.15T along

On a smooth plane inclined at 30°C with the horizontal, a thin current-carrying metallic rod is placed parallel to the horizontal ground. The plane is in a uniform magnetic field of 0.15T along the vertical direction. The value of the current in which the rod remain stationary is-  

Option: 1

12.3 A


Option: 2

11.3 A


Option: 3

10.3 A


Option: 4

14.3 A


Answers (1)

best_answer

Along the inclined plane, a component of the magnetic force acting on the conductor and the component of the weight of conduction will bring equilibrium so, 

B I L \cos \theta=m g \sin \theta \text {-(1) }

Here, force on the conductor BIL act horizontally towards right.

Now from equation 1 - 

I=\frac{m g \sin \theta}{B l \cos \theta}=\frac{m g}{B l} \tan \theta

I=\frac{0.30 \times 9.8}{0.15} \tan 30^{\circ}=11.3 \mathrm{~A}

Posted by

Gautam harsolia

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