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Two conductors carrying 1A current each ...

Two conductors carrying `1A` current each along east to west are held as shown in figure. The magnetic declination at the place is almost zero and angle of dip is `45^@`. The total intensity of earth's magnetic field at the place is `0*32` gauss. Calculate magnetic field at points `0*04m` below and above the two conductors.

Text Solution

Verified by Experts

Here, `I=1amp, r=0*04m`
Magnetic field due to two current carrying wires is `B=2xx(mu_0I)/(2pir)=(mu_0I)/(pir)`
`B=4xx10^-7xx(1)/(pixx0*04)=0*32xx10^-5T`
Earth's field, `R=0*32gauss =0*32xx10^-4T`
`H=Rcos delta=0*32xx10^-4cos45^@`
`=(0*32)/(sqrt2)xx10^-4T=0*226xx10^-4T`
`V=Rsindelta=0*32xx10^-4sin45^@`
`=(0*32)/(sqrt2)xx10^-4T=0*226xx10^-4T`
At a point above the wire, H and B are in same direction.

`:. R_H=B+H=0*32xx10^-5+0*226xx10^-4`
`=2*58xx10^-5T`
`:.` Resultant magnetic field `=sqrt(R_H^2+V^2)`
`=sqrt((2*58xx10^-5)^2+(0*226xx10^-4)^2)`
`=10^-5sqrt(11*7640)=3*429xx10^-5T`
At a point below the wire, H and B are in opposite direction.
`:.R_H^'=H-B=0*226xx10^-4-0*32xx10^-5m`
`=1*94xx10^-5T`
Resultant magnetic field `=sqrt(R_H^2+V^2)`
`=sqrt((1*94xx10^-5)^2+(2*26xx10^-5)^2)`
`=2*97xx10^-5T`
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