An ideal 200-turn solenoid having a lengthe of 19 cm and a distance of 2.4 cm ca

Question

An ideal 200-turn solenoid having a lengthe of 19 cm and a distance of 2.4 cm carries a current of 1.4 A. What is the magnitude mu of the solenoid's magnetic dipole moment? What is the magnitude of the magnetic field produced by the solenoid? A long wire carrying 0.91 A is placed 12 cm above the center of the solenoid such that the magnetic fields from the wire and the solenoid are pointed in opposite directions at the center of the solenoid. What is the magnitude of the net magnetic field at the center of the solenoid below the wire?

Explanation / Answer

a) Magnetic dipole moment, u = IA

         r = 2.4/2 = 1.2 cm = 0.012 m

A = pi r^2

u = 1.4 x (pi x 0.012^2) = 6.33 x 10^-4 A m^2

b) Magnetic field due to solenoid, B = u0 (N/L) I

B = 4 x pi x 10^-7 x (200 / 0.19) x 1.4 = 1.85 x 10^-3 T Or 1.85 mT

c) field due to wire , B' = u0 I / (2 pi d)

d = 12 cm = 0.12 m

B' = (4pi x 10^-7 x 0.91) / (2 x pi x 0.12) = 1.52 x 10^-6 T

Bnet = B - B' = Approax 1.85 x 10^-3 T

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