Question

A long solenoid has \(n = 1000\) turns/m and carries a current of \(I = 5\) A. Taking \(\mu_0 = 4\pi \times 10^{-7}\), what is the magnetic field inside the solenoid?

पर्याय

• \(B = 1.26 \times 10^{-2}\) T

• \(B = 6.28 \times 10^{-3}\) T

• \(B = 6.28 \times 10^{-6}\) T

• \(B = 2.00 \times 10^{-3}\) T

Answer 1

Using \(B = \mu_0 nI\): \(B = (4\pi \times 10^{-7}) \times 1000 \times 5 = 4\pi \times 10^{-7} \times 5000 = 6.28 \times 10^{-3}\) T. Thus, the magnetic field inside the solenoid is approximately \(6.28 \times 10^{-3}\) T (or about \(6.3 \times 10^{-3}\) T).