A long straight wire of radius a carries a steady current. The current is uniformly distributed across its cross-section.the ratio of the magnetic field at a4 inside and a4 outside from the surface of wire is:
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We know,magnetic field B in inside points = μ0ir2πR2 [where,R = radius,r = distance of the inside point from the axis and i = total current]
Distance from the axis at a4 inside = (a−a4) =3a4
So,Magnetic field at a4 inside = μ0×i×3a42πa2
Distance from the axis at a4 outside = (a+a4)
We also know,magnetic field B at outside points = μ0i2πr [r=distance of the outside point from the axis]
So,magnetic field at a4 outside = μ0i2π×5a4
The ratio of the magnetic field at a4 inside and a4 outside the surface of the wire = 3μ0i8πa2μ0i5πa
So,the correct option is C.
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A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a/2 and 2a is: