Question

# The magnetic field at the origin due to the current flowing in the wire is

A
μ0I8πa(^i+^k)
B
μ0I2πa(^i+^k)
C
μ0I8πa(^i+^k)
D
μ0I4πa2(^i^k)

Solution

## The correct option is A μ0I8πa(−^i+^k)One part of the wire is lying along positive x-axis, one part along the line x=z and one part along positive y-axis. There is no magnetic field at the origin due to the segments 1 and 2 (since they both pass through the origin). Hence, net magnetic field at origin is only due to 3. The magnetic field due to 3 at origin is given by the formula for the magnetic field due to a semi infinite wire at a point. B=μ0i4πd(sinα+sinβ) where α and β are the angles made by the line joining the point O with the extremes of the wire. Here, α=0∘,β=90∘ ∴B=μ0i4πd To find the direction of field, plot the magnetic field lines around the wire passing through point O. Then, we can say that the magnetic field at O must be perpendicular to the radius of the circle passing through O shown in the diagram. Since the radius is along the line x=z, the line perpendicular to it is x=−z i.e the direction of magnetic field at O is along (−^i+^k) Thus, magnetic field at O can be written as →B=μ0i4πdcos45∘(−^i)+μ0i4πdsin45∘(^k) =μ0i4π√2acos45∘(−^i+^k) =μ0i8πa(−^i+^k)

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