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» Here, `PQ=QR=r`. In art angled `DeltaPQR`, we have, `a^2=r^2+r^2=2r^2` or `r=a//sqrt2` Magnetic field at point P due to current through wire at Q is `B_1=(mu_0)/(4pi)(2I)/(r)=(mu_0I)/(2pir)=(mu_0I)/(2pi(a//sqrt2))=(mu_0I)/(sqrt2pia)` It is acting along PR. Magnetic field at point P due to current through wire at R is `B_2=(mu_0I)/(sqrt2pia)`, acting along PQ As `vecB_1` and `vecB_2` are acting perpendicular to each other, so the resultant magnetic field at P is `B=sqrt(B_1^2+B_2^2)=sqrt(((mu_0I)/(sqrtpia))^2+((mu_0I)/(sqrt2pia))^2)` `=sqrt(2)(mu_0I)/(sqrt2pia)=(mu_0I)/(pia)` It acts towards the mid-point of QR. |
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