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NEET PHYSICSMOVING CHARGES AND MAGNETISMEasy

Question

A long solenoid of radius 1 mm1 \text{ mm} has 100100 turns per mm. If 1 A1 \text{ A} current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:

A

6.28×10−4 T6.28 \times 10^{-4} \text{ T}

B

6.28×10−2 T6.28 \times 10^{-2} \text{ T}

C

12.56×10−2 T12.56 \times 10^{-2} \text{ T}

D

12.56×10−4 T12.56 \times 10^{-4} \text{ T}

Step-by-Step Solution

  1. Identify Formula: The magnetic field (BB) inside a long solenoid is given by B=μ0nIB = \mu_0 n I, where μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}, nn is the number of turns per unit length, and II is the current.
  2. Unit Conversion: The number of turns density is given in turns per mm. Convert this to SI units (turns per meter). n=100 turns/mm=100×103 turns/m=105 turns/mn = 100 \text{ turns/mm} = 100 \times 10^3 \text{ turns/m} = 10^5 \text{ turns/m}
  3. Substitute Values: μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7} \text{ T m/A} n=105 m−1n = 10^5 \text{ m}^{-1}
  • I=1 AI = 1 \text{ A} B=(4π×10−7)×(105)×1B = (4\pi \times 10^{-7}) \times (10^5) \times 1
  1. Calculate: B=4π×10−2 TB = 4\pi \times 10^{-2} \text{ T} B≈4×3.14×10−2 TB \approx 4 \times 3.14 \times 10^{-2} \text{ T} B=12.56×10−2 TB = 12.56 \times 10^{-2} \text{ T}

Exam Context & Concepts Covered

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

PHYSICSMOVING CHARGES AND MAGNETISMsolenoidradiuscurrentsolenoidmagnetic

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