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NEET PHYSICSELECTROMAGNETIC INDUCTIONEasy

Question

A square loop with a side length of 1 m1 \text{ m} and resistance of 1Ω1 \Omega is placed in a uniform magnetic field of 0.5 T0.5 \text{ T}. The plane of the loop is perpendicular to the direction of the magnetic field. The magnetic flux through the loop is:

A

zero

B

2 Wb2 \text{ Wb}

C

0.5 Wb0.5 \text{ Wb}

D

1 Wb1 \text{ Wb}

Step-by-Step Solution

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

  1. Identify Given Values: Magnetic Field (BB) = 0.5 T0.5 \text{ T}. Side length of square (ll) = 1 m1 \text{ m}. Area of loop (AA) = l2=(1 m)2=1 m2l^2 = (1 \text{ m})^2 = 1 \text{ m}^2. Resistance (RR) = 1Ω1 \Omega (This is distractor information not needed for flux calculation).

  2. Determine Angle (θ\theta):

  • The problem states the plane of the loop is perpendicular to the magnetic field. The area vector (A\mathbf{A}) is always normal (perpendicular) to the plane of the loop. Therefore, the angle θ\theta between the magnetic field B\mathbf{B} and the area vector A\mathbf{A} is 00^\circ.
  1. Calculate Flux: ΦB=(0.5 T)(1 m2)cos(0)\Phi_B = (0.5 \text{ T})(1 \text{ m}^2) \cos(0^\circ) ΦB=0.5×1×1=0.5 Wb\Phi_B = 0.5 \times 1 \times 1 = 0.5 \text{ Wb}

Thus, the magnetic flux through the loop is 0.5 Wb0.5 \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 INDUCTIONsquarelengthresistanceplaceduniform

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