JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Magnetic Effects of Current

As shown in the figure, a metallic rod of linear density

0.45 kg m^{- 1}

is lying horizontally on a smooth incline plane which makes an angle of

4 5^{\circ}

with 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 = 10 m / s^{2}

}

(A)30 A✓ Correct
(B)15 A
(C)10 A
(D)3 A

Answer: (A)

Solution

For equilibrium of the rod on the incline, the magnetic force component along the plane must balance the gravitational component along the plane. The magnetic force is

F = I L B

, where the effective component perpendicular to the rod is

I L B cos 4 5^{\circ}

. The gravitational component down the plane is

m g sin 4 5^{\circ}

. Setting these equal:

m g sin 4 5^{\circ} = I L B cos 4 5^{\circ}

. Solving for current:

I = \frac{m}{L} \cdot \frac{g}{B} = \frac{0.45 \times 10}{0.15} = 30

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Magnetic Effects of Current

Solution

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