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A solenoid that has length 30 cm, radius 1.2 cm, and 300 turns carries a current of 2.6 A. Find the magnitude of the magnetic field on the axis of the solenoid:

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Answer:

Step-by-step explanation:

length, l = 30 cm

Number of turns, N = 300

Current, i = 2.6 A

number of turns per unit length, n = N / l = 300 / 0.3 = 1000

The magnetic field due to the current carrying solenoid is


B=\mu _(0)ni

B = 4 x 3.14 x 10^-7 x 1000 x 2.6

B = 3.27 x 10^-3 T

B = 3.27 mT

User Nicolas Marshall
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