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NEET PHYSICSEasy

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}
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