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Question

A long wire carrying a current i is bent to form a plane angle α. Find the magnetic field B at point on the bisector of this angle situated at a distance x from the vertex.

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Solution

Given that,

Current =i

Angle =α

Magnetic field =B

Now, magnetic field at a point a from current carrying conductor making angle θ1 and θ2 with the ends of the wire is

B=μ0I4πa(sinθ1+sinθ2)

Now, a=xsinθ2

θ1=θ2

θ2=900

Now, the magnetic field

Blower=μ0I4π⎢ ⎢ ⎢ ⎢1+sin(θ2)xsin(θ2)⎥ ⎥ ⎥ ⎥

And Bupper is also same

No, if the current was flowing anticlockwise both these field would point upward outside the page

Now, net magnetic field

B=Blower+Bupper

B=2×μ0I4π⎢ ⎢ ⎢ ⎢1+sin(θ2)xsin(θ2)⎥ ⎥ ⎥ ⎥

B=μ0I2πx⎢ ⎢ ⎢ ⎢1+sin(θ2)sinθ2⎥ ⎥ ⎥ ⎥

B=μ0I2πx⎢ ⎢ ⎢ ⎢ ⎢{sin(θ4)+cos(θ4)}22sin(θ4)cos(θ4)⎥ ⎥ ⎥ ⎥ ⎥

Now, dividing numerator and denominator by cos2θ4

B=μ0I4πx⎢ ⎢ ⎢ ⎢ ⎢(1+tan(θ4))2tanθ4⎥ ⎥ ⎥ ⎥ ⎥

B=μ0I4πx[(1+tan(θ4))2]cot(θ4)

Now, the magnetic field is

B=Kcot(θ4)

Where, K=μ0I4πx[{1+tan(θ4)}2]

Hence, the magnetic field is B=Kcot(θ4)


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