As shown in the figure, a metallic rod of linear density 0.45 km-1 is lying horizontally on a smooth incline plane which makes an angle of 45° 5with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be:

{Use g = m/s2}

Option 1 - <p>30 A</p>
Option 2 - <p>15 A</p>
Option 3 - <p>10 A</p>
Option 4 - <p>3 A</p>
7 Views|Posted 8 months ago
Asked by Shiksha User
1 Answer
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8 months ago
Correct Option - 1
Detailed Solution:

Linear density, λ =  0.45 kg/m

Let length = l   = m = 0.45 l

B = 0.15 T

For equilibrium of rod :-

mg sin 45° = FB sin 45°

( 0 . 4 5 l ) g = l l B

So, l = 0 . 4 5 * 1 0 0 . 1 5 = 3 0 A

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Physics Moving Charges and Magnetism 2025

Physics Moving Charges and Magnetism 2025

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