A square coil has n = 31.8 turns per cm, side a = 6.96 cm, and length L = 32.9 c

Question

A square coil has n = 31.8 turns per cm, side a = 6.96 cm, and length L = 32.9 cm. If the current through this coil is I = 79.3 A, find the magnitude of the magnetic field in the center of the coil. Suppose now a second coil has the same dimensions and current, except it is long enough to be considered "infinite". In this case, find the magnitude of the magnetic field in the center of this coil. Calculate the percent difference between the magnitude of B in the first and second coils, percent difference = %

Explanation / Answer

given,

N = 31.8 turns

a = 6.96 cm = 0.0696 m

L = 32.9 cm = 0.329 m

a) magnetic fied at the center of square loop,

B1 = N*mue*I*a/(4*pi*sqrt(2)

= 31.8*4*pi*10^-7*79.3*0.0696/(4*pi*sqrt(2))

= 1.24*10^-5 T

b) magnetic field inside a solenoid, B2 = mue*N*I/L

= 4*pi*10^-7*31.8*0.329

= 1.315*10^-5 T

c)
(B1 - B2)*100/B1 = (1.24 - 1.315)*100/1.24

= -6.05 %

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