Q. 504.7( 3 Votes )

A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross section. Find the magnetic field B at a point at a distance.

(a) 2 cm

(b) 10 cm and

(c) 20 cm away from the axis. Sketch a graph of B versus x for 0 < x < 20 cm.

Answer :

Given:


Radius of wire(r) = 10 cm = 0.1 m


Current carried by it(i) = 5 A


Diagram:



Formula used:


Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ0 times the current enclosed by the surface.


,


Where B = magnetic field, dl = line element, μ0 = magnetic permeability of vacuum, I = current enclosed.


(a) Since the current is uniformly distributed over the cross section of the wire, at a distance of 2 cm (0.02 m) from the axis (inside the wire), I (current enclosed)


=


Where,


A = 0.2 A, where 0.1 m is the radius of the wire.


Hence, from Ampere’s circuital law,



where μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1, B = magnetic field, dl = line element, I = current enclosed


Substituting the values, we get


B x 2π x 0.02 = 4π x 10-7 x 0.2 (since 2π x 0.02 = circumference of the loop of radius 2 cm)


=> Magnetic field at a distance of 2 cm from the axis


= 2 x 10-6 T = 2μT (Ans)


(b) At a distance of 10 cm from the axis, we are basically on the surface of the wire. Hence the whole current of 5 A is enclosed by it.


Hence, from Ampere’s circuital ,,


where μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1, B = magnetic field, dl = line element, I = current enclosed


Here, I = 5 A.


Substituting the values, we get


B X 2π x 0.1 = 4π x 10-7 x 5(since 2π x 0.1 is the circumference of the loop of radius 10 cm)


=> Magnetic field at a distance of 10 cm from the axis


= 10-5 T = 10μT (Ans)


(c) At a distance of 20 cm from the axis, we are outside the wire. Hence the whole current of 5 A is enclosed by it.


Hence, from Ampere’s circuital law, ,


where μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1, B = magnetic field, dl = line element, I = current enclosed


Here, I = 5 A.


Substituting the values, we get


B X 2π x 0.2 = 4π x 10-7 x 5(since 2π x 0.2 is the circumference of the loop of radius 20 cm)


=> Magnetic field at a distance of 20 cm from the axis


= 5 x 10-6 T = 5μT (Ans)


Graph for B vs x for 0 < x < 20 cm:



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