Daily Olympiad: Physics - Magnetic Effects [20260511]

Challenge yourself with today's NEET practice! This test covers 'Magnetic Effects' for Physics (NEET - Dropper). Level: Hard | Duration: 45 mins.

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1. A circular coil of radius R carries a current I in a clockwise direction. An electron is moving with velocity v parallel to the axis of the coil at a point on the axis at distance z (z >> R) from the center. What is the direction of the magnetic force on the electron?

Solution
Correct: D
Using Biot-Savart law, the magnetic field at a large distance (z >> R) from the center of the coil along the axis can be approximated as a dipole field. The field direction at the axis of a clockwise current loop points into the page (using right-hand thumb rule). Since the electron is moving along the axis, its velocity is parallel to the magnetic field lines. The magnetic force on a charge is given by F = q(v × B). When the velocity and magnetic field are parallel, the cross product is zero. Hence, the force on the electron is zero.

2. Two infinite parallel wires separated by a distance 2d carry currents I in opposite directions. A square loop of side d is placed between them, centered at a distance d from each wire. What is the net torque on the loop in the magnetic field produced by the two wires?

Solution
Correct: A
The magnetic field from each wire at the position of the square loop cancels due to symmetry. The top and bottom arms of the loop experience equal and opposite forces, while the left and right arms also cancel each other's forces since the loop is equidistant from both wires. Hence, all forces cancel, resulting in zero net torque on the loop.

3. A rectangular loop PQRS is moving with velocity v perpendicular to a uniform magnetic field B. The loop has fixed dimensions and moves in a plane perpendicular to the field. The side PQ of length L is fixed, and side RS is a sliding conductor that moves with velocity v. What is the induced EMF in the loop as RS moves outward by a small displacement dx?

Solution
Correct: B
The induced EMF in a sliding conductor is given by Faraday's law, EMF = B * v * L. The displacement dx is not directly part of the EMF expression since it's the rate of change of flux. The flux change is dΦ = B * dA/dt = B * L * v. Hence, the induced EMF is directly BLv, independent of dx.

4. A copper disc of radius R rotates with angular velocity ω in a uniform magnetic field B perpendicular to its plane. A potential difference develops between the axis and the rim of the disc. What is the magnitude of this potential difference?

Solution
Correct: A
This is a case of motional EMF in a rotating disc. Each radial element of the disc moves perpendicular to the magnetic field. The EMF is given by integrating dE = Bvdr from 0 to R. Since v = ωr, we get E = ∫₀ᴿ Bωr dr = ½BωR². The potential difference between axis and rim is this value.

5. Two concentric coplanar circular loops of radii a and 3a carry currents I₁ and I₂ respectively. What is the condition for the magnetic induction at the center of the loops to be zero?

Solution
Correct: D
The magnetic field at the center of a loop is μ₀I/(2R). For concentric loops with radii a and 3a, the fields oppose each other if currents are in opposite directions. For net zero field, μ₀I₁/(2a) = μ₀I₂/(2*(3a)). Solving gives I₁/I₂ = 1/3. Reversing direction for one loop to cancel the other satisfies the condition.