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A long, straight, cylindrical wire of radius R carries a current uniformly distributed over its cross section. At what location is the magnetic field produced by this current equal to half of its largest value? inside the iwre

Answer :

Answer:

[tex]r=\dfrac{R}{2}[/tex]

Explanation:

We know that maximum value of magnetic field in the long wire

[tex]B_{max}=\dfrac{\mu _oI}{2\pi R}[/tex]

I=Current  ,R=Radius of wire  ,B= magnetic field

μo=Constant

At distance r the magnetic filed is the half of the maximum magnetic filed

At distance r

[tex]B=\dfrac{\mu _oIr}{2\pi R^2}[/tex]

[tex]B=\dfrac{B_{max}}{2}[/tex]

So we can say that

[tex]\dfrac{\mu _oIr}{2\pi R^2}=\dfrac{1}{2}\times \dfrac{\mu _oI}{2\pi R}[/tex]

[tex]r=\dfrac{R}{2}[/tex]

Therefore the answer is [tex]r=\dfrac{R}{2}[/tex]

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