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  • 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:  <img alt="(\mu_0 = 4\pi \times 10^{-7} T m A^{-1})" src="http://entrancecorner.codecogs.com/gif.latex?%28%5Cmu_0%20%

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:  (\mu_0 = 4\pi \times 10^{-7} T m A^{-1})
Option: 1 6.28 \times 10^{-4}T
Option: 2 3.14 \times 10^{-4}T
Option: 3 6.28 \times 10^{-5}T
Option: 4 3.14 \times 10^{-5}T

Answers (2)

best_answer

\begin{aligned} &\text { Given: } \quad L=50 \mathrm{cm}=0.5 \mathrm{m} \quad \mathrm{i}=2.5 \mathrm{A} \quad \mathrm{N}=100 \text { turns }\\ &\text { So, number of turns per unit length } \mathrm{n}=\frac{\mathrm{N}}{\mathrm{L}}=\frac{100}{0.5}=200 \mathrm{turns} / \mathrm{m}\\ &\text { Magnetic field at the centre of the solenoid } \quad \mathrm{B}=\mu_{\mathrm{o}} \mathrm{ni}=4 \pi \times 10^{-7} \times 200 \times 2.5=6.28 \times 10^{-4} \mathrm{T} \end{aligned}

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Deependra Verma

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Guru G

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