Question

How much conventional current must you run in a solenoid with radius = 0.05 m and length = 0.39 m to produce a magnetic field inside the solenoid of 2 × 10−5 T, the approximate field of the Earth? The solenoid has 200 turns.

Answer 1

Answer:

Explanation:

radius of the solenoid, r = 0.05 m

length of the solenoid, l = 0.39 m

Magnetic field of the solenoid, B = 2 x 10^-5 T

Number of turns, N = 200

The magnetic field of the solenoid is given by

$B=\mu _{0}ni$

where, i be the current and n be the number of turns per unit length

n = N / l = 200 / 0.39 = 512.8

$2\times 10^{-5}=4 \times 3.14\times 10^{-7}\times 512.8\times i$

i = 0.031 A