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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Apllication of Ampere's law (I), Application of Ampere's law (II) are considered the most difficult concepts.

  • 33 Questions around this concept.

Solve by difficulty

The negatively and uniformly charged nonconducting disc as shown in the figure is rotated clockwise with great angular speed. The direction of the magnetic field at point A in the plane of the disc is

A current i ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at  a/2 and 2a is:

A current I flows along the length of an infinitely long, straight, thin walled pipe. Then

A horizontal overhead powerline is at a height of 4 m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is

(\mu _{0}=4\pi\times 10^{-7}\; T\: m\: A^{-1} )

Concepts Covered - 3

Ampere's circuital law

Ampere's circuital law

Like Biot-savart law, Amperes law is also a method to calculate the magnetic field due to a given current distribution.

So, it says that 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. 

    \begin{array}{l}{\text { i.e. } \oint \vec{B} d \vec{l}=\mu_{0} \sum i=\mu_{0}\left(i_{1}+i_{3}-i_{2}\right)} \\ \\ { \text { Also using }\left.\vec{B}=\mu_{0} \vec{H} \text { \ \ \ \ \ . . . . .where } \vec{H}=\text { magnetising field }\right)} \\ \\ {\oint \mu_{0} \vec{H} \cdot \overrightarrow{d l}=\mu_{0} \Sigma i \Rightarrow \oint \vec{H} \cdot \overrightarrow{d l}=\Sigma i}\end{array}    

 

Now, We can see that the total current crossing the above area is (i_1+i_3-i_2), so any current outside the given area will not be consider. So we have assume  

 

Apllication of Ampere's law (I)

Apllication of Ampere's law -

Magnetic field due to infinite straight wire - 

As by Biot-savart 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}.\vec{dl} = \mu_o.i_{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 Cos \theta = 1

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

                                                                                          B\oint dl = \mu_o.i

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

                                                                                           \\ B.2 \pi d = \mu_o i \\ \\ B = \frac{\mu_o i}{2 \pi d}

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

Application of Ampere's law (II)

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

Let us consider a 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 \overrightarrow{B_{\mathrm{in}}} \cdot d \vec{l}=\mu_{0} i

 

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

Then from the top it can be seen as - 

                                                                       

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} 

 

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

                            \text { Hence at inside point } \oint \overrightarrow{B_{\mathrm{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 - 

                                                                

 

Study it with Videos

Ampere's circuital law
Apllication of Ampere's law (I)
Application of Ampere's law (II)

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