A coil of wire 7 m long and with 220 turns carries a current of 1.6 A.
Part(1)B=μ0NI2πrPart(1)\\B=\frac{\mu_0 NI}{2\pi r}Part(1)B=2πrμ0NI
B=4π×220×1.62π×7B=\frac{4\pi \times220\times1.6}{2\pi \times7}B=2π×74π×220×1.6
B=1.0057×10−5TB=1.0057\times10^{-5}TB=1.0057×10−5T
Part(2)
B=μ0H(1+χ)B=\mu_0H(1+\chi)B=μ0H(1+χ)
H=NilH=\frac{Ni}{l}H=lNi
H=220×1.67=50.28A/mH=\frac{220\times1.6}{7}=50.28A/mH=7220×1.6=50.28A/m
χ=μμ0−1\chi=\frac{\mu}{\mu_0}-1χ=μ0μ−1
Put value
B=6.320×10−6TB=6.320\times10^{-6}TB=6.320×10−6T
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