Ampere's Circuital Law And Its Applications - Practice Questions & MCQ

Ampere's circuital law

Amperes law is also a method to calculate the magnetic field due to a given current distribution like Biot-Savart's law.

Statement: The line integral of the magnetic field $\vec{B}$ around any closed curve is equal to $\mu_o$ times the total current i passing through the area enclosed by the curve.
Mathematical statement: $\oint \vec{B} d \vec{l}=\mu_0 \sum i=\mu_0\left(i_1+i_3-i_2\right)$

Also using $\vec{B}=\mu_0 \vec{H} \quad \ldots . .($ where $\vec{H}=$ magnetising field)

$
\oint \mu_0 \vec{H} \cdot \overrightarrow{d t}=\mu_0 \Sigma_i \Rightarrow \oint \vec{H} \cdot \overrightarrow{d t}=\Sigma i
$
 

Fingers are curled in the loop direction, the current in the direction of the thumb is taken as positive whereas in the direction opposite to that of the thumb is taken as negative.
Now, we can see that the total current crossing the above area is $\left(i_1+i_3-i_2\right)$, so any current outside the given area will not be considered. So we have to assume ( Outward $\odot \rightarrow+v e$, Inward $\otimes \rightarrow-v e$ )

General guidelines for the selection of Ampere's path for its application in different situations

(i) Path should be chosen in such a way that at every point of the path magnetic induction should be either tangential to the path elements or normal to it so that the 'dot' product can be easily handled.

(ii) Path should be chosen in such a way that at every point of the path magnetic induction should either be uniform or zero so that calculations become easy.

Application of Ampere's law -

Magnetic field due to infinite straight wire -
As by Biot-savant law, we have calculated the magnetic field. Now with the help of Ampere's circuital law, let us take an infinite straight wire, now if we want to calculate the magnetic field of this wire at a distance of 'd' from the wire. Then,

$
\oint \vec{B} \cdot \overrightarrow{d l}=\mu_o \cdot i_{\text {inside }}
$
 

See the figure given below, it represents the magnetic field produced by the current-carrying wire and the length vector. We can see that the 'B' and 'dl' is making angle 0 degree, so the $C$ os $\theta=1$

Also B is constant as the current is constant which is flowing through the wire. So we can write that -

$
B \oint d l=\mu_o \cdot i
$

Now the length of the loop is the circumference of the loop. Here the radius is 'd' and full integration of 'dl' is equal to $2 \pi d$. So we can write that -

$
\begin{aligned}
& B .2 \pi d=\mu_o i \\
& B=\frac{\mu_o i}{2 \pi d}
\end{aligned}
$

One can notice that the result obtained is the same as that we have obtained by Biot-savant law.

Magnetic field due to infinitely long cylindrical wire due to current 'i' flowing through its surface only - 

Let us consider an infinitely long cylindrical wire of radius R and the current is distributed on the surface of the wire, then this wire will behave as a hollow cylindrical wire.

Now let us take different situations -                                   

a) For a point inside the wire - (r<R)

From the top view, the Ampere's loop will look like this - 

Since there is no current inside the Ampere's loop, so there will be no magnetic field in this loop because -

$
\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 i
$
 

b) For a point outside the wire (r>R) -

Then from the top, it can be seen as - 

$
B=\frac{\mu_o i}{2 \pi r}
$

c) On the surface $(r=R)$

$
B_s=\frac{\mu_o i}{2 \pi r}
$

From the above equations, we can plot a graph between B and different positions ' r '.

Magnetic field due to infinitely long cylindrical wire due to current 'i' distributed uniformly across its cross-section - 

Magnetic field due to a cylindrical wire is obtained by the application of Ampere's law. Here also we consider few cases one by one - 

a) Outside the cylinder -

It is just like the concept of a current carrying wire which we have studied in the last concept with the help of Ampere's circuital law as well as by Biot-savart law. So again by applying same Ampere's circuital law we can deduce that -

$
B=\frac{\mu_o i}{2 \pi r}
$
 

b) Inside the solid cylinder : Current enclosed by loop (i') is lesser then the total current (i) - 

Since the current density will remain same. So,

$
J=J^{\prime} \Rightarrow \quad i=i \times \frac{A^{\prime}}{A}=i\left(\frac{r^2}{R^2}\right)
$

Hence at inside point $\oint \overrightarrow{B_{\text {in }}} \cdot d \vec{l}=\mu_0 i \Rightarrow B=\frac{\mu_0}{2 \pi} \cdot \frac{i r}{R^2}$
c) At surface $(r=R)$ -

$
B_s=\frac{\mu_o i}{2 \pi r}
$
 

The variation of B with r can be drawn as -