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Two straight parallel conductors are 50c...

Two straight parallel conductors are `50cm` apart and carry oppositively directed currents, held perpendicular to the plane of paper as shown in figure. The current in first conductor, `I_1=20A` and in the second conductor is `I_2=24A`. A point P is separated from the first conductor by a distance of `40cm` and the second conductor by `30cm`. Calculate the magnetic field at point P.

Text Solution

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In, figure, `CD=50cm`, `CP=40cm` and `DP=30cm`

As `CD^2=CP^2+DP^2`
i.e., `50^@=40^@+30^@`, so `/_CPD=90^@`
Magnetic field induction at P due to downward current `I_1` through wire C will be
`B_1=(mu_0)/(4pi)(2I_1)/(r_1)=10^-7xx(2xx20)/(0*40)`
`=10^-5T` along P.D.
Magnetic field induction at P due to upward current `I_2` through wire at D will be
`B_2=(mu_0)/(4pi)(2I_2)/(r_2)=10^-7xx(2xx24)/(0*30)`
`=1*6xx10^-5T` along P.C.
Resultant magnetic field induction at P
`B=sqrt(B_1^2+B_2^2)=sqrt((10^-5)^2+(1*6xx10^-5)^2)`
`=1*89xx10^-5T`
Let `alpha` be the angle which the resultant magnetic field induction `vecB` make with the direction of `vecB_2`, i.e., `/_BPC=alpha`. Then
`tan alpha=B_1/B_2=(10^-5)/(1*6xx10^-5)=10/16=0*625=tan32^@`
`:. alpha=32^@`, with the line joining the point P with the first conductor.
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