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GUJCET · Physics · Moving Charges and Magnetism
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 250 turns. It carries a current of 5 A . What is the magnitude of the magnetic field inside the solenoid ?
- A \(3.14 \times 10^{-3} T\)
- B \(6.28 \times 10^{-3} T\)
- C \(62.8 \times 10^{-3} T\)
- D Zero
Answer & Solution
Correct Answer
(A) \(3.14 \times 10^{-3} T\)
Step-by-step Solution
Detailed explanation
A
Magnetic field inside the solenoid is,
\(\begin{array}{l} B =\mu_0 n I \\ B =\frac{\mu_0 NI }{l} \\ B=\frac{4 \pi \times 10^{-7} \times 250 \times 5}{0.5} \\ B=31400 \times 10^{-7} T \\ B =3.14 \times 10^{-3} T\end{array}\)
Magnetic field inside the solenoid is,
\(\begin{array}{l} B =\mu_0 n I \\ B =\frac{\mu_0 NI }{l} \\ B=\frac{4 \pi \times 10^{-7} \times 250 \times 5}{0.5} \\ B=31400 \times 10^{-7} T \\ B =3.14 \times 10^{-3} T\end{array}\)
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