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NEET PHYSICSMAGNETISM AND MATTEREasy

Question

Two identical bar magnets are fixed with their centres at a distance dd apart. A stationary charge QQ is placed at PP in between the gap of the two magnets at a distance DD from the centre OO as shown in the figure. The force on the charge QQ is:

A

zero.

B

directed along OPOP.

C

directed along POPO.

D

directed perpendicular to the plane of the paper.

Step-by-Step Solution

  1. Lorentz Force Law: The magnetic force (F\mathbf{F}) exerted on a charge qq moving with velocity v\mathbf{v} in a magnetic field B\mathbf{B} is defined by the equation: F=q(v×B)\mathbf{F} = q(\mathbf{v} \times \mathbf{B}) [Source 191, Eq. 4.3]
  2. Condition of the Charge: The problem specifies that the charge QQ is stationary. This means its velocity vector v=0\mathbf{v} = 0.
  3. Force Calculation: Substituting v=0\mathbf{v} = 0 into the Lorentz force equation: F=Q(0×B)=0\mathbf{F} = Q(0 \times \mathbf{B}) = 0 The cross product of a zero vector with any field vector is zero.
  4. Conclusion: A stationary electric charge does not experience any force due to a magnetic field, regardless of the strength or direction of the field produced by the bar magnets .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MAGNETISM AND MATTER. 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.

PHYSICSMAGNETISM AND MATTERidenticalmagnetscentresdistancestationary

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