Magnetic field at the center of a circular current-carrying loop of N turns and radius r is
B=Nμ₀I/2r….. (1)
N=L/2πr ….(2)
I=U/R=U/R₀L ….(3)
Substitution (2) and (3) in (1) gives
B=Nμ₀I/2r = (L/2πr) •(μ₀/2r) •(U/R₀L) = μ₀U/4πr²R₀=
=4π•10⁻⁷•12/ 4π •(0.2)² •5.5•10⁻³ =5.45•10⁻³ T
A piece of copper wire has a resistance per unit length of 5.50 x 10^-3 Ω/m. The wire is wound into a thin, flat coil of many turns that has a radius of 0.200 m. The ends of the wire are connected to a 12.0-V battery. Find the magnetic field strength at the center of the coil.
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1 answer