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NEET PHYSICSELECTROMAGNETIC INDUCTIONEasy

Question

A circular disc of the radius 0.2 m0.2 \text{ m} is placed in a uniform magnetic field of induction 1π Wb m2\frac{1}{\pi} \text{ Wb m}^{-2} in such a way that its axis makes an angle of 6060^\circ with B\vec{B}. The magnetic flux linked to the disc will be:

A

0.02 Wb

B

0.06 Wb

C

0.08 Wb

D

0.01 Wb

Step-by-Step Solution

The magnetic flux (ΦB\Phi_B) through a surface is defined as the scalar product of the magnetic field (B\vec{B}) and the area vector (A\vec{A}): ΦB=BA=BAcosθ\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta

  1. Identify parameters: Magnetic Field (BB) = 1π Wb m2\frac{1}{\pi} \text{ Wb m}^{-2} Radius (rr) = 0.2 m0.2 \text{ m} Area (AA) = πr2=π(0.2)2=0.04π m2\pi r^2 = \pi (0.2)^2 = 0.04\pi \text{ m}^2 Angle (θ\theta): The area vector A\vec{A} is directed along the normal to the surface, which coincides with the axis of the disc. The problem states the axis makes an angle of 6060^\circ with B\vec{B}. Therefore, θ=60\theta = 60^\circ .

  2. Calculation: Substituting the values into the flux formula: ΦB=(1π)(0.04π)cos(60)\Phi_B = \left( \frac{1}{\pi} \right) (0.04\pi) \cos(60^\circ) ΦB=0.04×0.5\Phi_B = 0.04 \times 0.5 ΦB=0.02 Wb\Phi_B = 0.02 \text{ Wb}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROMAGNETIC INDUCTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROMAGNETIC INDUCTIONcircularradiusplaceduniformmagnetic

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