Question

A 2.0-cm-diameter, 15-cm-long solenoid is tightly wound with one layer of wire. A 2.5 A current through the wire generates a 3.0 mT magnetic field inside the solenoid. What is the diameter of the wire, in mm?

Answer 1

Answer:

d = 1.047 mm

Explanation:

given,

diameter of the wire = 2.0-cm

length of solenoid = 15 cm = 0.15

Current in the wire = I = 2.5 A

magnetic field = B = 3.0 mT

Magnetic field inside the solenoid

        $B = \dfrac{\mu_0 N I}{L}$

        $B = \dfrac{\mu_0 N I}{L}$

               N x d = l

        $N = \dfrac{l}{d}$

        $B = \dfrac{\mu_0 \dfrac{l}{d} I}{L}$

        $3 \times 10^{-3} = \dfrac{4\pi \times 10^{-7}\times \dfrac{0.15}{d}\times 2.5}{0.15}$

        $0.45 \times 10^{-3}\ d = 4\pi \times 10^{-7}\times 0.15\times 2.5$

        $\ d = \dfrac{4\pi \times 10^{-7}\times 0.15\times 2.5}{0.45 \times 10^{-3}}$

               d = 1.047 x 10⁻³ m

               d = 1.047 mm

diameter of the wire is d = 1.047 mm