Question

The current density inside a long, solid, cylindrical wire of radius a = 5.0 mm is in the direction of the central axis and its magnitude varies linearly with radial distance r from the axis according to J = J0r/a, where J0 = 420 A/m2. Find the magnitude of the magnetic field at a distance (a) r=0, (b) r = 2.9 mm and (c) r=5.0 mm from the center.

Answer 1

Step-by-step explanation:

Given that,

Radius a= 5.0 mm

Radial distance r= 0, 2.9 mm, 5.0 mm

Current density at the center of the wire is given by

`J_(0)=420\ A/m^2`

Given relation between current density and radial distance

`J=(J_(0)r)/(a)`

We know that,

When the current passing through the wire changes with radial distance,

then the magnetic field is induced in the wire.

The induced magnetic field is

`B=(\mu_(0)i_(ind))/(2\pi r)`...(I)

We need to calculate the induced current

Using formula of induced current

`i_(ind)=\int_(0)^(r){J(r)dA}`

`i_(ind)= \int_(0)^(r){(J_(0)r)/(a)2\pi r}`

`i_(ind)={(2\pi J_(0))/(a)}\int_(0)^(r){r^2}`

`i_(ind)={(2\pi J_(0))/(a)}{(r^3)/(3)}`

We need to calculate the magnetic field

Put the value of induced current in equation (I)

`B=\frac{\mu_(0){(2\pi J_(0))/(a)}{(r^3)/(3)}}{2\pi r}`

`B=(\mu_(0)J_(0)r^2)/(3a)`

(a). The magnetic field at a distance r = 0

Put the value into the formula

`B=(4\pi*10^(-7)*420*0)/(3*5.0*10^(-3))`

`B = 0`

The magnetic field at a distance 0 is zero.

(b). The magnetic field at a distance r = 2.9 mm

`B=(4\pi*10^(-7)*420*(2.9*10^(-3))^2)/(3*5.0*10^(-3))`

`B = 2.95*10^(-7)\ T`

The magnetic field at a distance 2.9 mm is
`2.95*10^(-7)\ T`

(c). The magnetic field at a distance r = 5.0 mm

`B=(4\pi*10^(-7)*420*(5.0*10^(-3))^2)/(3*5.0*10^(-3))`

`B = 8.79*10^(-7)\ T`

The magnetic field at a distance 5.0 mm is
`8.79*10^(-7)\ T`

Hence, This is the required solution.