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Physics · Moving Charges and Magnetic Field

NEET (UG) 2024 — Question 33

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A . The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π×10−7SI4 \pi \times 10^{-7} \mathrm{SI} units):

  1. Option A:

    44 mT

  2. Option B:

    4.4 T

  3. Option C:

    4.4 mT

    Correct
  4. Option D:

    44 T

Answer: C

Step-by-step solution

The magnitude of magnetic field due to circular coil of NN turns is given by

BC=μ0iN2RB_{C}=\frac{\mu_{0} i N}{2 R}

=4π×10−7×7×1002×0.1=\frac{4 \pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1}

=4.4×10−3 T=4.4 \times 10^{-3} \mathrm{~T}

=4.4mT=4.4 \mathrm{mT}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2024
Paper
2024 paper
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law

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