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Repeat Exercise 28.43 for the case in which the current in the central, solid conductor is , the current in the tube is , and these currents are in the same direction rather than in opposite directions.

Short Answer

Expert verified

(a) The magnetic field at points outside the central, solid conductor but at inside the tube isB=μ0I12πr .

(b) The magnetic field at points outside the central, solid conductor but at outside the tube isB=μ0I1+I22πr .

Step by step solution

01

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behaving like a magnet.

02

Determine the magnetic field at points outside the central, solid conductor but at inside the tube and at points outside the central, solid conductor but at outside the tube

a)

Using ampere’s law to a circular path of radius:

Bdl=BIBdl=B(2πr)

But the enclosed current isIencl=I1

Bdl=μ0lendB(2πr)=μ0l1B=μ0l12πr

Hence, the magnetic field at points outside the central, solid conductor but at inside the tube isB=μ0l12πr. .

b)

Now the magnetic field at points outside the central, solid conductor but outside the tube is calculated as:

The enclosed current is the sum of two currents, so,

Iend=I1+I2

So, the magnetic field can be calculated as:

B=μ0I1+I22πr

Hence, the magnetic field at points outside the central, solid conductor but at outside the tube isB=μ0I1+I22πr .

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