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KCET · Physics · Magnetic Effects of Current

A solenoid of length \(50 \mathrm{~cm}\) having 100 turns carries a current of \(2.5 \mathrm{~A}\). The magnetic field at one end of the solenoid is

  1. A \(6.28 \times 10^{-4} \mathrm{~T}\)
  2. B \(1.57 \times 10^{-4} \mathrm{~T}\)
  3. C \(9.42 \times 10^{-4} \mathrm{~T}\)
  4. D \(3.14 \times 10^{-4} \mathrm{~T}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3.14 \times 10^{-4} \mathrm{~T}\)

Step-by-step Solution

Detailed explanation

Given, length of solenoid, \(l=50 \mathrm{~cm}=0.5 \mathrm{~m}\)
Number of turns, \(N=100\)
Current, \(I=2.5 \mathrm{~A}\)
Number of turns per unit length in the solenoid
\(n=\frac{N}{l}=\frac{100}{0.5}=200 \text { turns } / \mathrm{m}\)
Magnetic field at the one end of solenoid,
\(\begin{aligned}B & =\frac{\mu_0 n I}{2}=\frac{4 \pi \times 10^{-7} \times 200 \times 2.5}{2} \\& =10 \pi \times 10^{-5}=\pi \times 10^{-4}=3.14 \times 10^{-4} \mathrm{~T}\end{aligned}\)