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 in side the solenoid of 2 × 10−5 T, the approximate field of the Earth? The solenoid has 200 turns.
1 answer:
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
where, i be the current and n be the number of turns per unit length
n = N / l = 200 / 0.39 = 512.8
i = 0.031 A
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