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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×10−74\pi \times 10^{-7} SI units)

A

4.4 T

B

4.4 mT

C

44 T

D

44 mT

Step-by-Step Solution

  1. Identify Formula: The magnitude of the magnetic field (BB) at the center of a circular coil carrying current is given by the formula: B=μ0NI2RB = \frac{\mu_0 N I}{2R} where: μ0\mu_0 is the permeability of free space (4π×10−7 T m/A4\pi \times 10^{-7} \text{ T m/A}) NN is the number of turns II is the current RR is the radius
  2. Substitute Values: N=100N = 100 I=7 AI = 7 \text{ A}
  • R=10 cm=0.1 mR = 10 \text{ cm} = 0.1 \text{ m} B=(4π×10−7)×100×72×0.1B = \frac{(4\pi \times 10^{-7}) \times 100 \times 7}{2 \times 0.1}
  1. Calculation: Use π≈227\pi \approx \frac{22}{7} for simpler calculation: B=4×227×10−7×100×70.2B = \frac{4 \times \frac{22}{7} \times 10^{-7} \times 100 \times 7}{0.2} B=4×22×10−50.2B = \frac{4 \times 22 \times 10^{-5}}{0.2} B=88×10−50.2B = \frac{88 \times 10^{-5}}{0.2} B=440×10−5 TB = 440 \times 10^{-5} \text{ T} B=4.4×10−3 TB = 4.4 \times 10^{-3} \text{ T}
  2. Convert Units: Since 10−3 T=1 mT10^{-3} \text{ T} = 1 \text{ mT}: B=4.4 mTB = 4.4 \text{ mT}
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