“The world is a
dangerous place, not because of those who do evil, but because of those who
look on and do nothing.”

– Albert Einstein.

Today we will discuss two questions pertaining to
the magnetic field produced by infinitely long current carrying cylinders. The first
question is the usual single correct answer type multiple choice question. The
second one is an integer answer type question, the answer to which is a single
digit integer ranging from 0 to 9. Here are the questions with their solution.

(1) An infinitely long conducting cylinder with
inner radius

*R*/2 and outer radius*R*carries a uniform current density along its length. The magnitude of the magnetic field**B**as a function of the radial distance*r*from the axis is best represented by**Case (i)**:

*r*<

*R*/2

Using Ampere’s circuital law we find that the
magnetic fiel at points at distance

*r*<*R*/2 is zero.
[We have ∫

**B**.**d****ℓ***=*μ_{0}*I*.
Since the current passing through the surface
enclosed by the path of integration is

*zero*, the value of**B**is zero at distance*r*< R/2]**Case (ii)**

For distance

*r*lying between*R/*2 and R (or,*R/*2 ≤*r*< R) we have
∫

**B**.**d****ℓ***=*μ_{0}[π*r*^{2}J*– π(**R/*2)^{2}J] where J is the current density
[π

*r*^{2}J*– π(**R/*2)^{2}J is the current passing through the surface enclosed by the path of integration in this case]
Or,

*B*×2π*r =*μ_{0}[π*r*^{2}J*– π(**R/*2)^{2}J]
This gives

*B =*(μ_{0}J/2)[*r*– (*R*^{2}/4*r*)]**Case (iii)**

For

*r*≥*R*we have**B**.

**d**

**ℓ**

*=*μ

_{0}[π

*R*

^{2}J

*– π(*

*R/*2)

^{2}J]

Or,

*B*×2π*r =*μ_{0}[π*R*^{2}J*– π(**R/*2)^{2}J]
This gives

*B =*(μ_{0}J/2*r*)[*R*^{2}*– (**R*^{2}/4*r*)] = 3μ_{0}J*R*^{2}/8*r*
The graph shown in option (D) indicates the above
three cases correctly.

(2) A cylindrical cavity of
diameter ‘a’ exists inside a cylinder of diameter 2a as shown in the figure.
Both the cylinder and the cavity are infinitely long. A uniform current density

*J flows along the length. If the magnitude of the magnetic field at the point P is given by N μ*_{0 }aJ/12, then the value of N is
If
the cylinder has no cavity, the magnetic flux density

**B**at the point P is given by_{1}
∫

**B**._{1}**d****ℓ***=*μ_{0}πa^{2}J
Therefore,

*B*_{1}×2πa = μ_{0}πa^{2}J
Or,

*B*_{1}= μ_{0}aJ/2
Since there exists a cavity, the current is
reduced by π(a/2)

^{2}J and the magnetic field is*reduced*by**B**._{2}
The field

**B**_{2}_{ }is given by
∫

**B**._{2}**d****ℓ***=*μ_{0}π(a/2)^{2}J
Therefore,

*B*_{2}×2π(3a/2) = μ_{0}π(a/2)^{2}J
[The circular path of integration in this case
has radius a+(a/2) = 3a/2]

Or,

*B*_{2}= μ_{0}aJ/12
The magnitude

*B*of the magnetic field at the point P due to the actual conductor with the cavity is given by*B = B*

_{1}–

*B*

_{2}= μ

_{0}aJ/2– μ

_{0}aJ/12

Or, B = 5μ

_{0}aJ/12
In the question the above field is given as Nμ

_{0}aJ/12
Therefore

**N = 5**.