explain with steps pleaseSolutionWe will use the Amperes law

explain with steps please?

Solution

We will use the Ampere\'s law to find expression for the magnetic field at varying distance from the axis.

For x < a,

Let us consider a loop of radius x which is concentric with the axis, we will apply Ampere\'s law around this loop to get the magnetic field.

Here, the current density is I/a², hence the current enclosed within the loop selected would be I x²/ a²

Using Ampere\'s law, we can write: _o I x²/ a² = B(2x)

or, B = _oIx/ 2a²

For a < x < b

Again, we consider a loop of radius x such that a < x < b and concentric with the axis. Here the curent enclosed would be I, hence the ampere\'s law would give us:

B(2x) = _o I

or, B = _o I / 2x is the required expression for magnetic field.

For b < x < c

Here again, we will have to use the current density of the outer shell to calculate the current enclosed.

Current density of the outer shell would be: I / (c² - b²)

So the ampere\'s law gives us:B(2x) = _o [I - I(x² - b²) / (c² - b²)] = _o I [(c² - x²) / (c² - b²)] [Since the direction of the currents is opposite we need to subtract one from another]

or, B = _o I (c² - x²) / 2x(c² - b²) is the required expression for the magnetic field.

For x > c

Here, the net current enclosed would be zero, hence magnitude of magnetic field would be zero throughout.

Therfore we get the magnetic field as:

B = _oIx/ 2a² For x < a,

B = _o I / 2x For a < x < b

B = _o I (c² - x²) / 2x(c² - b²) For b < x < c

B = zero For x > c