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NEET PhysicsMedium

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : (μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1})

1

6.28 \times 10^{-4} \text{ T}

2

3.14 \times 10^{-4} \text{ T}

3

6.28 \times 10^{-5} \text{ T}

4

3.14 \times 10^{-5} \text{ T}

Step-by-Step Solution

Magnetic field at centre of solenoid = μ0nI\mu_0 n I. n=NL=10050×10−2=200 turns/mn = \frac{N}{L} = \frac{100}{50 \times 10^{-2}} = 200 \text{ turns/m}. I=2.5 AI = 2.5 \text{ A}. On putting the values, B=4π×10−7×200×2.5=6.28×10−4 TB = 4\pi \times 10^{-7} \times 200 \times 2.5 = 6.28 \times 10^{-4} \text{ T}.

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