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The current density J inside a long, solid, cylindrical wire of radius a=3.1mm is in the direction of the central axis, and its magnitude varies linearly with radial distance rfrom the axis according toJ=J0r/a, where J0=310A/m2. (a) Find the magnitude of the magnetic field at role="math" localid="1663132348934" r=0, (b) Find the magnitude of the magnetic fieldr=a/2 , and(c) Find the magnitude of the magnetic field r=a.

Short Answer

Expert verified
  1. The magnitude of the magnetic field atr=0is zero.
  2. The magnitude of the magnetic field atr=a2isB=0.10μT
  3. The magnitude of the magnetic field at r=a is B=0.40μT

Step by step solution

01

Listing the given quantities

  • The radius of the wirea=3.1mm=0.0031m
  • Current density varies asJ=J0ra
  • J0=310A/m2
02

Understanding the concept of magnetic field and Ampere’s law

The relation between current and current density is,

i=JdA

Ampere’s law states that,

Bds=μ0i

The line integral in this equation is evaluated around a closed-loop called an Amperian loop.The current ion the right side is the net current encircled by the loop.

By using the current density equation in Ampere’s law and integrating it with respect to distance r, we can get the general equation for the magnetic field due to the current-carrying cylindrical wire. By using this, we can find the value of the magnetic field at given distances.

03

Explanation

According to Ampere’s law,

Bds=μ0iencBds=μ0ienc

Sinceds=lengthofthecircularpath=2πr

B(2πr)=μ0ienc (i)

Current is given by,

ienc=JdA

Using the given current density J=J0ra

ienc=J0radA

Area of a differential element of the circle isdA=2πrdr

ienc=J0ra2πrdr=2πJ0ar2dr=2πJ0ar33

Using this in equation (i),

B(2πr)=μ02πJ0ar33

Therefore,

B=μ0J03r2a (ii)

04

(a) Calculations of the magnitude of magnetic field at r=0

At r=0, Equation (ii) becomes,

B=μ0J03×0a=0

Thus, the magnetic field at r=0 is zero.

05

(b) Calculations of the magnitude of magnetic field at r=a/2

Atr=a2,Equation2becomes

B=μ0J03×(a2)2aB=μ0J0a12=4π×10-7×310×0.003112=4×3.14×10-7×310×0.003112=0.1×10-6T=0.10μT

Thus, The magnetic field at r=a2 is B=0.10μT

06

(c) Calculations of the magnitude of magnetic field at r=a 

Atr=a,Equation(2)becomes

B=μ0J03×a2a=μ0J0a3=4π×10-7×310×0.00313=4×3.14×10-7×310×0.00313=4.02×10-7T=0.40μT

Thus, the magnetic field at r=a is B=0.40μT.

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