Different Rules and Formulas Relating the Sides and Angles of the Triangle: Rules, Formulas

May 20, 202239 Insightful Publication

**Amperes Circuital Law**: The Amperes Circuital law is used to calculate the magnetic field due to distributed currents. It is analogous to Gauss law in Electrostatics, which calculates the electric field due to distributed charges. We have studied the continuous distribution and discrete distribution of charges and masses. Ampere Circuital Law formula gives a method to calculate the magnetic field due to a given current distribution.

According to Gauss law in Electrostatics, “total electric flux through a closed surface enclosing a charge is \(\frac{1}{{{\varepsilon _0}}}\) times the magnitude of the charge enclosed, i.e., \(\phi = \frac{1}{{{\varepsilon_0}}}\left( {{Q_{enc}}} \right)\). In this article, we will understand the concept and statement of Ampere’s Circuital Law, its applications, and its demerits.

**Study Magnetic Field In A Solenoid**

- Ampere’s Circuital Law was proposed by French Physicist
**AndrÃ©-Marie Ampere**. - Ampere was born on January 20, 1775, in Lyon, France. He was schooled at home by his father, and he showed an aptitude for mathematics at an early age. Ampere was a child prodigy, he never went to school, read widely, and had an excellent memory.
- Ampere was a Mathematician, Physicist who is best known for the study of Electrodynamics, Ampere’s Law, confirmation and amplification of the work of Oersted on the relationship of electricity and magnetism, one who invented the astatic needle, which is a critical component of the modern astatic galvanometer.
- He was the first to demonstrate that when two parallel wires are charged with electricity, a magnetic field is generated.
- He is generally credited as one of the first to discover electromagnetism.
- The unit of electric current ‘ampere’ is named after him.

“**The line integral of the magnetic field \(\overrightarrow B \) around any closed curve is equal to \({\mu _0}\) times the net current \(i\) threading through the area enclosed by the curve**“.

i.e. \(\oint {\mkern 1mu} {\mkern 1mu} \overrightarrow B \overrightarrow {dl} = {\mu _0}\sum i = {\mu _0}\left( {{i_1} + {i_3} – {i_2}} \right)\)

where

\(\mu_0 =\) permeability of free space

\(\overrightarrow B =\) magnetic field at a point on the boundary of the surface making an angle ‘\(\theta\)’ with the length element ‘\(\overrightarrow {dl}\)’.

\(\oint {\mkern 1mu} {\mkern 1mu} \overrightarrow B \overrightarrow {dl} =\) The sum of all the \(\overrightarrow B \cdot \overrightarrow {dl} \) products over the complete loop is also called the ‘Amperian loop.’

**Note:**

1. Total current crossing the above loop is \(({i_1} + {i_3} – {i_2})\). Any current outside the area is not included in net current, but while calulating \(\oint \overrightarrow B \cdot \overrightarrow {dl} \), we have to include magnetic field due to all currents (both inside as well as outside the loop currents)

2. Sign convention : (Outward current \(\to\) positive, Inward current \(\to\) negative)

3. When the direction of the current is away from the observer, the closed path’s direction is clockwise. When the direction of current is towards the observer, the direction of the closed path is anticlockwise.

4. This law is valid only for steady currents. This law holds well irrespective of the size and shape of the closed path (Amperian loop) enclosing the current.

5. The statement \(\oint {\overrightarrow B \overrightarrow {dl} } = 0\) does not necessarily mean that the magnetic field \(\overrightarrow B\) is zero everywhere along the path but that no net current is passing through the path.

We know that

\(\oint {\overrightarrow B \overrightarrow {dl} = {\mu _0}\sum i } = {\mu _0}({i_1} + {i_3} – {i_2})\)

By using \(\overrightarrow B = {\mu _0}\overrightarrow H \) (where \(H =\) magnetizing field)

\(\oint {\overrightarrow B \overrightarrow {dl} = {\mu _0}\sum i } = {\mu _0}({i_1} + {i_3} – {i_2})\)

\( = \oint {{\mu _0}\overrightarrow H .\overrightarrow {dl} } = {\mu _0}\Sigma i\)

\( \Rightarrow \oint {\overrightarrow H .\overrightarrow {dl} } = \Sigma i\).

James Clerk Maxwell explained that Ampere’s Law is valid only for steady current or when the electric field does not change with time. To see this inconsistency consider a parallel plate capacitor being charged by a battery. During charging, time-varying current flows through connecting wires.

Applying Ampere’s Law for loop \(l_1\) and \(l_2\)

For loop \(1 -\) \(\oint_{{l_1}} {\overrightarrow B \cdot \overrightarrow {dl} } = {\mu _0}i\)

For loop \(2 -\) \(\oint_{{l_2}} {\overrightarrow B \cdot \overrightarrow {dl} } = 0\) (\(i = 0\) between the plates).

But practically, it is observed that there is a magnetic field between the plates when the plate is getting charged or discharged. Hence Ampere’s Law fails,

i.e., \(\oint_{{l_1}} {\overrightarrow B \cdot \overrightarrow {dl} } \ne {\mu _0}i\).

Maxwell assumed that some current must be flowing between the capacitor plates during the charging process. He named it displacement current. Hence modified law is as follows:

\(\oint {\overrightarrow B \cdot \overrightarrow {dl} } = {\mu _0}({i_c} + {i_d})\) or \(\oint {\overrightarrow B } \cdot \overrightarrow {dl} = {\mu _0}({i_c} + {\varepsilon _0}\frac{{d{\phi _E}}}{{dt}})\)

where

\({i_c} =\) Conduction current \(=\) current due to flow of charges in a conductor and

\({i_d} =\) Displacement current \( = {\varepsilon _0}\frac{{d{\phi _E}}}{{dt}} = \) current due to the changing electric field between the plates of the capacitor.**Note:**

1. Displacement current magnitude \((i_d) =\) conduction current magnitude \((i_c)\).

2. \(i_c\) and \(i_d\) in a circuit, may not be continuous, but their sum is always continuous.

Ampere’s law is used

- To find the magnetic field due to a cylindrical wire.
- To find the magnetic field due to an infinite sheet carrying current.
- To find the magnetic field inside a solenoid and a toroid.
- To find the magnetic field inside a conductor.
- To find forces between current-carrying conductors.

Consider an infinitely long conducting wire carrying a current \(I\). Let \(P\) be a point close to it at a distance \(r\) from it (see figure (a)). Consider a circular path of radius \(r\) passing through the point \(P\). Let \(\overrightarrow B \) be the magnetic field at \(P\) due to the conductor (see figure (b)).

Since all the points on the circular path are at the same distance, \(r\), \(B\) is the same at all points. Let \(\overrightarrow {dl}\) be a small element on the circular path. The magnetic field \(\overrightarrow {B}\) is along the tangent to the circle of radius \(r\).

From Ampere’s Circuital Law,

\(\oint {\overrightarrow B } \cdot \overrightarrow {dl} = {\mu _0}I\)

Since \(\overrightarrow B\) and \(\overrightarrow {dl}\) are parallel

\(\oint \overrightarrow B \cdot \overrightarrow {dl} = \oint B\;dl = B\left( {2\pi r} \right)\)

\(\therefore \,B(2\pi r) = {\mu _0}I\)

\( \Rightarrow B = \frac{{{\mu _0}I}}{{2\pi r}}\)

The variation of the magnetic field \(\overrightarrow B\) with distance \(r\) is shown in the figure:

In all the above cases magnetic field outside the wire at \(P\) is \(\oint {\overrightarrow B \cdot \overrightarrow {dl} = {\mu _0}i} \Rightarrow B\int {dl} = {\mu _0}i \Rightarrow B \times 2\pi r = {\mu _0}i \Rightarrow {B_{out}} = \frac{{{\mu _0}i}}{{2\pi r}}\).

We know that \({B_{out}} = \frac{{{\mu _0}i}}{{2\pi r}}\). On the surface of the cylinder, \(r = R\).

Therefore \({B_{{\text{surface}}}} = \frac{{{\mu _0}i}}{{2\pi R}}\)

If an Amperian loop is drawn inside a hollow cylinder, the loop doesn’t enclose any current, but in a solid cylinder, the Amperian loop encloses a finite current.

Therefore, the magnetic field inside the hollow cylinder is zero, whereas the magnetic field inside the solid cylinder is not zero.

**Note:**

1. For all cylindrical current distributions

\({B_{\text{axis}}} = 0(\min )\), \({B_{{\text{surface}}}} = (\max )\) (distance \(r\) always from the axis of cylinder),

\({B_{{\text{out}}}} \propto \frac{1}{r}\).

The figure shows an infinite sheet of current with linear current density \( \rm{j(A/m)}\). Due to symmetry, the field line pattern above and below the sheet is uniform. Consider a square loop of side \(l\) as shown in the figure.

According to Ampere’s Law, \(\int_a^b {B \cdot dl} + \int_b^c {B \cdot dl} + \int_c^d {B \cdot dl} + \int_d^a {B \cdot dl} = {\mu _0}i\)

Since \(B \bot dl\) along the path \(b \to c\) and \(d \to a\), therefore, \(\int_b^c {B.dl} = 0\); \(\int_d^a {B.dl} = 0\)

Also, \(B\parallel dl\) along the path \(a \to b\) and \(c \to d\), thus \(\int_a^b {B \cdot dl} + \int_d^a {B \cdot dl} = 2Bl\)

The current enclosed by the loop is \(i = jl\)

Therefore, according to Ampere’s Law \(2Bl = {\mu _0}(jl)\) or \(B = \frac{{{\mu _0}j}}{2}\).

A cylindrical coil of many tightly wound turns of insulated wire with a coil diameter smaller than its length is called a solenoid. One solenoid end behaves like the North Pole, and the opposite end behaves like the South Pole. As the length of the solenoid increases, the interior field becomes more uniform, and the external field becomes weaker.

A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of the solenoid.

If \(N =\) total number of turns,

\(l =\) length of the solenoid

\(n =\) number of turns per unit length \( = \frac{N}{l}\)

The magnetic field inside the solenoid at point \(P\) is given by

\(B = \frac{{{\mu _0}}}{{4\pi }}(2\pi \,ni)[\sin \alpha + \sin \beta ]\).

If the solenoid has infinite length and the point is inside the solenoid, i.e. \(\alpha = \beta = (\pi /2)\).

Then, \({B_{in}} = {\mu _0}ni\)

If the solenoid is of infinite length and the point is near one end, i.e. \(\alpha = 0\) and \(\beta = \left( {\frac{\pi }{2}} \right)\)

Then, \({B_{end}} = \frac{1}{2}\left( {{\mu _0}ni} \right)\)**Note:**

1. The magnetic field outside the solenoid is zero.

2. \({B_{end}} = \frac{1}{2}{B_{in}}\)

A toroid can be considered a ring-shaped closed solenoid. Hence it is like an endless cylindrical solenoid.

Consider a toroid having n turns per unit length.

Let \(i\) be the current flowing through the toroid (figure). The magnetic lines of force mainly remain in the core of the toroid and are in the form of concentric circles. Consider such a circle of mean radius \(r\). The circular closed path surrounds \(N\) loops of wire, each of which carries current \(i\), therefore, from \(\oint {\overrightarrow B \cdot \overrightarrow {dl} = {\mu_0}{i_{net}}} \)

\( \Rightarrow B \times (2\pi r) = {\mu _0}Ni\) \( \Rightarrow B = \frac{{{\mu _0}Ni}}{{2\pi r}} = {\mu _o}ni\) where \(n = \frac{N}{{2\pi r}}\).**Note:**

1. For any point inside the space surrounded by toroid and outside the toroid, magnetic field \(B\) is zero because the net current enclosed in these spaces is zero.

*Q.1. Derive an expression for Magnetic field intensity inside a uniform solid cylinder and a uniform thick hollow cylinder.Ans:*

Solid cylinder | Inside the thick portion of a hollow cylinder |

Current enclosed by loop \(i\)’ is lesser, then the total current \(i\) Current density is uniform, i.e. \(J = j’ \Rightarrow \frac{i}{A} = \frac{{i’}}{{A’}}\) \( \Rightarrow i’ = i \times \frac{{A’}}{A} = i\,\left( {\frac{{{r^2}}}{{{R^2}}}} \right)\) Hence at point \(Q\) \(\oint {\overrightarrow B \cdot d\overrightarrow l = {\mu _0}i’} \) \( \Rightarrow B \times 2\pi r = \frac{{{\mu _0}i{r^2}}}{{{R^2}}}\) \( \Rightarrow B = \frac{{{\mu _0}}}{{2\pi }}.\frac{{ir}}{{{R^2}}}\) | Current enclosed by loop \(i\)’ is lesser, then the total current \(i\) Also \(i’ = i \times \frac{{A’}}{A} = i \times \frac{{({r^2} – R_1^2)}}{{(R_2^2 – R_1^2)}}\) Hence at point \(Q\) \(\oint {\overrightarrow B \cdot d\overrightarrow l = {\mu _0}i’} \) \( \Rightarrow B \times 2\pi r = {\mu _0}i \times \frac{{({r^2} – R_1^2)}}{{(R_2^2 – R_1^2)}}\) \( \Rightarrow B = \frac{{{\mu _0}i}}{{2\pi r}}.\frac{{({r^2} – R_1^2)}}{{(R_2^2 – R_1^2)}}\). If \(r = R_1\) (inner surface), \(B = 0\) If \(r = R_2\) (outer surface) \(B = \frac{{{\mu _0}i}}{{2\pi {R_2}}}\) (max.) |

** Q.2. Derive an expression for a Magnetic field inside a solenoid of infinite length.** Let \(N =\) total number of turns,

Ans:

\(l =\) length of the solenoid

\(n =\) number of turns per unit length \(= \frac {N}{l}\)

\(I =\) current flowing through the solenoid

To use Ampere’s Circuital Law to determine the magnetic field inside a solenoid, we choose a rectangular closed path \(PQRSP\) where \(PQ = l\).

The line integral of \(B\) over the closed path \(PQRSP\) is given by:

\(\int {\overrightarrow B } \cdot \overrightarrow {dl} = \int\limits_P^Q {\overrightarrow B } \cdot \overrightarrow {dl} + \int\limits_R^S {\overrightarrow B } \cdot \overrightarrow {dl} + \int\limits_S^P {\overrightarrow B } \cdot \overrightarrow {dl} \)

Now \(\int\limits_P^Q {\overrightarrow B } \cdot \overrightarrow {dl} = \int\limits_p^Q {Bdl} \cos {0^0} = B\int\limits_p^Q {dl} = Bl\)

Also \(\int\limits_Q^R {\overrightarrow B \cdot \overrightarrow {dl} } = \int\limits_S^P {\overrightarrow B \cdot \overrightarrow {dl} = 0} \) (Since the angle between \(\overrightarrow B\) and \(\overrightarrow {dl}\) is \(90^\circ\)

And \(\int\limits_R^S {\overrightarrow B \cdot \overrightarrow {dl} } = 0\) (Since field outside the solenoid is zero)

\(\therefore \,\int {\overrightarrow B \cdot \overrightarrow {dl} = Bl} \) ———(i)

According to Ampere’s Circuital Law,

\(\oint {\overrightarrow B \cdot d\overrightarrow l = {\mu _0}i’} \)

Here \(i\)^{‘} (current enclosed) by the rectangle \(PQRS\) is \(NI\).

\(Bl = {\mu _0}NI\)

From equations (i) and (ii),

\(Bl = {\mu _0}NI\)

\( \Rightarrow {B_{in}} = {\mu _0}ni\) (where \(n = \frac {N}{l}\)).

** Q.3. A long solenoid has \(200\) turns per cm and carries a current of \(2.5\;\rm{A}\). What is the magnetic field at its centre?** \(B = {\mu _0}ni = 4\pi \times {10^{ – 7}} \times \frac{{200}}{{{{10}^{ – 2}}}} \times 2.5 = 6.28 \times {10^{ – 2}}\,{\text{Wb/}}{{\text{m}}^2}\).

Ans:

** Q.4. The average radius of a toroid made on a ring of non-magnetic material is \(0.1\;\rm{m}\), and it has \(500\) turns. If it carries \(0.5\;\rm{A}\) current, what is the magnetic field produced along its circular axis inside the toroid?** \(B = {\mu _0}ni\); where \(n = \frac{N}{{2\pi R}}\)

Ans:

\(\therefore \,B = 4\pi \times {10^{ – 7}} \times \frac{{500}}{{2\pi \times 0.1}} \times 0.5\)

\( = 5 \times {10^{ – 4}}\,{\text{T}}\).

*Q.5. For the solenoid shown in the figure, what is the magnetic field at point \(P\)?*

** Ans:** \(B = \frac{{{\mu _0}}}{{4\pi }}.2\pi \,ni\,(\sin \alpha + \sin \beta )\)

From figure \(\alpha = \left( {{{90}^o} – {{30}^o}} \right) = {60^o}\) and \(\beta = \left( {{{90}^o} – {{60}^o}} \right) = {30^o}\)

\(\therefore \,B = \frac{{{\mu _0}ni}}{2}(\sin {60^o} + \sin {30^o}) = \frac{{{\mu _0}ni}}{4}(\sqrt 3 + 1)\).

*Q.6. The figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius \(R\) and outer radius \(2R\), a cylinder is carrying uniformly distributed current along its axis. What will be the magnetic induction at point \(P\) at a distance \(\frac {3R}{2}\) from the cylinder’s axis?*

** Ans:** By using \(B = \frac{{{\mu _0}i}}{{2\pi \,r}}\left( {\frac{{{r^2} – {a^2}}}{{{b^2} – {a^2}}}} \right)\), here \(r = \frac{{3R}}{2},\) \(a = R\), \(b = 2R\)

\(B = \frac{{{\mu _0}i}}{{2\pi \left( {\frac{{3R}}{2}} \right)}} \times \left[ {\frac{{\left( {\frac{{3R}}{2}} \right) – {R^2}}}{{{{\left( {2R} \right)}^2} – {R^2}}}} \right] = \frac{{5 \cdot {\mu _0}i}}{{36\pi r}}\)

\( = \frac{{5.{\mu _o}i}}{{36\pi \,r}}\).

- Ampere’s Circuital Law states that “The line integral of the magnetic field \(\overrightarrow B\) around any closed curve is equal to \(\mu_0\) times the net current ‘\(i\)’ threading through the area enclosed by the curve.”
- Ampere’s Circuital Law is analogous to Gauss law in Electrostatics.
- Ampere’s Circuital Law becomes invalid when the electric current is not steady.

**Magnetic Field on the Axis of a Circular Current Loop**

Check the most frequently asked questions about Ampere’s Critical Law.

**Q.1: Name the scientist who performed experiments with forces that act on current-carrying wires?**

**Ans: **AndrÃ©-Marie Ampere

**Q.2: State true or false: Ampere’s Circuital Law and Magnetic field due to a uniform ring concept is used to determine the magnetic field inside an infinite solenoid.**

**Ans:** True.

**Q.3: State Ampere’s circuital law.**

**Ans:** Ampere’s circuital law states that “The line integral of the magnetic field around any closed curve is equal to times the net current threading through the area enclosed by the curve.”

**Q.4: State true or false: If the direction of the current is reversed, the direction of the magnetic field reverses.**

**Ans:** True.

**Q.5: What is the magnetic field intensity at a point on the axis of an infinitely long, straight, thin-walled tube carrying a current \(I\)?**

**Ans:** Zero because the tube is hollow from inside, and the Amperian loop doesn’t subtend any current.

**Q.6: Is Ampere’s Circuital Law universal?**

**Ans: **No, because Ampere’s Circuital Law is valid only when the current flowing is steady.

**Q.7: Which Physicist removed the inconsistency in Ampere’s Circuital Law?**

**Ans:** James Clerk Maxwell.

**Q.8: What are the applications of Ampere’s Circuital Law?**

**Ans:** Ampere’s Law is used

1. To find the magnetic field due to a cylindrical wire.

2. To find the magnetic field due to an infinite sheet carrying current.

3. To find the magnetic field inside a solenoid and a toroid.

4. To find the magnetic field inside a conductor.

5. To find forces between current-carrying conductors.

**Q.9: \(\oint {\overrightarrow B } \overrightarrow {dl} = 0\), Does it necessarily mean that everywhere along the path?**

**Ans: **The statement \(\oint {\overrightarrow B } \overrightarrow {dl} = 0\) does not necessarily mean \(\overrightarrow B = 0\) everywhere along the path but that no net current is passing through the path.

**We hope this detailed article on Ampere’s circuital law helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel free to ask us in the comment section and we will be more than happy to assist you. Happy learning!**

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