A solenoid of length 1.2 m and radius of 2.0 cm has 300 turns and a core of relative permeability 500. It carries a current of 3.5 A. The magnitude of the magnetic field inside the solenoid is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.55 T
Inside a long solenoid the field is uniform, \( B = \mu_0 n I \) for an air core. A core of relative permeability \( \mu_r \) multiplies this by \( \mu_r \). The radius does not enter the formula; only the number of turns per unit length does.
\( \displaystyle n = \frac{N}{l} = \frac{300}{1.2\,\mathrm{m}} = 250\,\mathrm{m^{-1}} \)
\( \displaystyle B = \mu_r \mu_0 n I \)
\( \displaystyle B = 500 \times 4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \times 250\,\mathrm{m^{-1}} \times 3.5\,\mathrm{A} \)
\( \displaystyle B \approx 0.55\,\mathrm{T} \)