InterviewSolution
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InterviewSolution
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Figure shows a right-angled isosceles `DeltaPQR` having its base equal to a. A current of I ampere is passing downwards along a thin straight wire cutting the plane of paper normally as shown at Q. Likewise a similar wire carries an equal current passing normally upwards at R. Find the magnitude and direction of the magnetic field induction B at P. Assume the wires to be infinity long. |
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Answer» Given `PQ=r_1`, `PR=r_2` In `DeltaPQR`, `a^2=r_1^2+r_2^2` Magnetic field induction at point P due to the current passing through wire at Q is `B_1=(mu_0I)/(2pir_1)` acting along PR Magnetic field induction at point P due to current passing through wire at R is `B_2=(mu_0I)/(2pir_2)` acting along PQ. Since `vecB_1` and `vecB_2` are perpendicular to each other, the resultant magnetic field induction at P is `B=sqrt(B_1^2+B_2^2)=(mu_0I)/(2pi)[(1/r_1)^2+(1/r_2)^2]^(1//2)` `=(mu_0I)/(2pi)[(r_1^2+r_2^2)/(r_1^2r_2^2)]^(1/2)=(mu_0I)/(2pi)((a^2)/(r_1^2r_2^2))^(1/2)=(mu_0Ia)/(2pir_1r_2)` |
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