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NEET PHYSICSMedium

A long solenoid of radius 1 m1\text{ m} has 100 turns per mm100\text{ turns per mm}. If 1 A1\text{ A} current flows in the solenoid, the magnetic field strength at the centre of the solenoid is

1

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

2

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

3

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

4

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

Step-by-Step Solution

Magnetic field at the center of a long solenoid is B=μ0nIB = \mu_0 n I. Here 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}. I=1 AI = 1\text{ A}. B=4π×10−7×105×1=4π×10−2≈12.56×10−2 TB = 4\pi \times 10^{-7} \times 10^5 \times 1 = 4\pi \times 10^{-2} \approx 12.56 \times 10^{-2}\text{ T}.

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