Learn Ampere's Law with problems & examples, practice to find the correct Amperian loop for long straight wire having steady current and calculate the magnetic field in a solenoid of length.

- Consider a current carrying wire, as shown in figure.
- Using right hand rule, the magnetic field will be created around a current carrying wire in circular loop.

- The tangent at any point on a circular loop gives the direction of magnetic field at that point due to current carrying wire.
- If integration of magnetic field is performed around all the points of circle, the initial and final point around the integration path will be same.
- So, the line integral around the closed curve is denoted by \(\oint B\cdot d\ell\)

- According to Ampere's law, the line integral of magnetic field around any closed path is given as

\(\oint\vec B\cdot d\vec \ell =\mu_0\;I_{enclosed}\)

- For integrating in clockwise direction, use right hand thumb rule to get the positive direction.

- Curl your fingers around the arrow, the direction of thumb gives the positive direction of current.

**Case 1 :**For clockwise integration

\(I_{enclosed}=I_3-I_1-I_2\)

**Case 2 : **For anticlockwise integration

\(I_{enclosed}=I_1+I_2-I_3\)

**NOTE **

According to Ampere's Law

\(\oint\vec B\cdot d \vec \ell=\mu_0 I_{enclosed}\)

where \(I_{enclosed}\) includes all the current enclosed in loop and \(\vec B\) is not due to \(I_{enclosed}\) only, in fact it is due to all current carrying elements inside(enclosed) or outside. \(\vec B\) is the property of space.

A For Ampere's Law, \(\oint\vec B \cdot d \vec \ell=\mu_0\;I_{enclosed}\) where \(I_{enclosed}=I_3-I_1-I_2\)

B \(\vec B\) is only due to \(I_1\) , \(I_2\) and \(I_3\)

C \(\vec B\) is only due to \(I_1 ,I_2,I_3\) and \(I_4\)

D None of these

- The graphical representation of magnetic field is represented by certain lines known as magnetic field lines.

- Point the right hand thumb in the direction of current.
- Curl the fingers around the wire to indicate circle.
- Fingers point in the direction of magnetic field around the wire.

- A long wire wound in the form of helix with negligible distance between turns, forms a solenoid.
- When a current flows in a solenoid of finite length such that the turns of solenoid are closely spaced, it produces magnetic field lines similar to bar magnet.

- When the length of solenoid becomes greater than the space of turns, exterior field becomes weaker and the interior becomes more uniform.

- Consider an infinite sheet of current placed in Y-Z plane such that the current is in y-direction (out of paper).
- Magnetic field lines will be in loop such that these are perpendicular to the direction of current, as shown in figure.

- For a cylinder carrying current, magnetic field lines are in circle around the surface such that current pierces those circles.

- Amperian loop is same as Gaussian surface or loop of any shape but for easier calculation of magnetic field, Amperian loop is chosen as follows:

- The loop is chosen such that \(d\vec\ell\) is either perpendicular or parallel to the field.
- For \(d\vec\ell\) parallel to the field, the field must be constant over the loop.

- For a current carrying wire, the Amperian loop will be a circular loop whose \(d\vec\ell\) and magnetic field vector \((\vec B)\) are parallel to each other.

\(\vec B\parallel d\vec \ell\)

- Consider loop-2, this loop encloses a small current as the charges in a wire move coil by coil along the solenoid length. Hence, it produces a very weak field inside a solenoid. So, this selection of amperian loop for solenoid is incorrect.
- Consider loop-1 of rectangular shape of length \(\ell\) and width \(w\). The magnetic field \(\vec B\) is uniform inside the ideal solenoid.

** For Side 3**

- Both magnetic field vector and \(d\vec\ell\) are perpendicular. So, the contribution along this side is zero. \(\vec B\perp d\vec \ell\)

** For Side 2 and Side 4**

- For both sides, the magnetic field vector and \(d\vec \ell\) are perpendicular to each other. So, the contribution along these sides is also zero.

\(\vec B\perp d\vec \ell\)

** For side 1**

- Since, magnetic field vector and \(d\vec \ell\) are parallel to each other and also \(\vec B\) is uniform along this side so, this choice of loop-1 as Amperian loop for solenoid is correct.

\(\vec B\parallel d\vec s,\;\vec B\) is uniform.

- Consider an infinite long conductor having current \(I\).
- A point P situated at a distance \(x\) from the rod.
- To calculate magnetic field, we know for an infinite long conductor any point outside the conductor can be considered as mid-point.

- Taking Amperian loop around point P.

\(\oint B\cdot dx=\mu_0I\)

\(\Rightarrow\;B\oint dx=\mu_0I\)

\(\Rightarrow\;B×2\pi\,x=\mu_0I\)

\(B=\dfrac {\mu_0I}{2\pi\,x}\)

A \(6×10^{–4}\) \((-\hat k)T\)

B \(3 ×10^{–6}\) \(\hat k\;T\)

C \(0.8 ×10^{–7}\) \((-\hat k)\;T\)

D \(4 ×10^{–5}\) \(\hat k\;T\)

- For ideal solenoid, the strength of magnetic field is uniform and strong inside it and weaker at the exterior of solenoid.

- Considering two loops for ideal solenoid, as shown in figure.

- Since loop 2 carries very small current and the field is weaker. So, this choice of Amperian loop is not correct.
- For loop 1, considering side 1, the magnetic field is uniform and both \(d\vec \ell\) and \(\vec B\) are parallel. So, this choice of Amperian loop is correct.

- Taking integration over the closed rectangular path

\(\oint\vec B\cdot d\vec s= \displaystyle\int\limits_{\text {path 1}} \vec B\cdot d\vec s+ \displaystyle\int\limits_{\text {path 2}} \vec B\cdot d\vec s+ \displaystyle\int\limits_{\text {path 3}} \vec B\cdot d\vec s+ \displaystyle\int\limits_{\text {path 4}} \vec B\cdot d\vec s\)

Since path 3 is outside the solenoid, where \(\vec B\) is taken as zero so,

\( \displaystyle\int\limits_{\text { path 3}}\vec B\cdot d\vec s=0\)

\(= \displaystyle B\int\limits_{\text {path 1}} ds\,cos0°+ \displaystyle\int\limits_{\text {path 2}} B\, ds\; cos 90°+ \displaystyle\int\limits_{\text {path 3}} B\, ds\; cos 90° + \displaystyle\int\limits_{\text {path 4}} B\, ds\; cos 90°\)

\(\oint\vec B\cdot d\vec s= B\ell+0+0+0=B\ell \) \(\Big\{ \therefore \int ds=\ell\)

- Total current through the rectangular path equals the current through each turn multiplied by the number of turns.
- If N is the number of turns in the length \(\ell\), then total current through the rectangle is \(NI\).

\(I_{\text{(enclosed)}}=NI\)

\(\oint \vec B \cdot d\vec s=\mu_0\,I_{\text {(enclosed)}}\)

\(B\ell=\mu_0NI\)

\(B=\dfrac {\mu_0NI}{\ell}\)

\(B=\mu_0nI\)

where \(n=\dfrac {N}{\ell}\)= number of turns per unit length.

A \(15× 10^{–3}\, T\)

B \(2\pi× 10^{–3}\, T\)

C \(44\ T\)

D \(16× 10^{–3}\, T\)

- Consider a long cylinder of radius R carrying a steady current \(I\) that is uniformly distributed over the cross-sectional area, as shown in figure.

**Case 1 : **Magnetic field in a region outside the cylinder \((r\geq R)\)

- To calculate magnetic field in a region outside the cylinder, choose Amperian loop of radius r such that \((r\geq R)\), as shown in figure.

- By symmetry, \(\vec B\) is constant in magnitude and parallel to \(d\vec s\) at each point on this circle (Amperian loop).

By applying Ampere's Law

\(\oint \vec B\cdot d \vec s=B\oint ds\)

\(\Rightarrow B×2\pi\;r=\mu_0I\)

\(B=\dfrac {\mu_0I}{2\pi\,r}\) (for \(r \geq R\))

**Case 2 **: Magnetic Field in a Region inside the Cylinder (\(r<R\))

- To calculate magnetic field in a region inside the cylinder, choose Amperian loop such that r < R , as shown in figure.

- Let \(I'\) be the current passing through cross-sectional area \(\pi r^2\)

\(\dfrac {I'}{I}=\dfrac {\pi r^2}{\pi R^2}\)

\(I'=\left ( \dfrac {r^2}{R^2} \right)I\)

- From Ampere's Law

\(\oint\vec B \cdot d \vec s=B(2\pi\,r)=\mu_0I'\)

\(B=\dfrac {\mu_0}{2\pi\,r} \left ( \dfrac {r^2}{R^2} \right)I\) \(\left [I'=\left (\dfrac {r^2}{R^2}\right)I\right]\)

\(B=\left ( \dfrac {\mu_0I}{2\pi\,R^2} \right)r\) for (r < R )

A B1 = 48 × 10 –7 T, B2 = 10 –7 T

B B1 = 48 × 10 –5 T, B2 = 10 –7 T

C B1 = 24 × 10 –7 T, B2 = 10 –6 T

D B1 = 48 × 10 –7 T, B2 = 10 –5 T

- Consider a thick sheet of current having current density \(\vec J\) outside the page is confined in x-z plane.
- To calculate the magnetic field at point x above sheet, consider that the sheet is being made up of large number of parallel wires, all carrying current in the same direction.

- Magnetic field due to wires at the left of origin.

- Magnetic field due to wires at the right of origin.

- The y component of magnetic field (B
_{y}) due to wires at the left of origin cancel the y component of magnetic field (B_{y}) due to wires at the right of origin. - Thus, only x-component of magnetic field B
_{x}is left. - Therefore, at any point on y-axis above the origin (y >0)

? \(B=-B \hat i\)

and at any point on y-axis below the origin (y < 0)

\(B=B \hat i\)

- Consider an Amperian loop of rectangular path of length L.

From Ampere's Law

\(\oint \vec B \cdot d\vec \ell=\mu_0\,I_{\text {(enclosed)}}\)

\(\displaystyle \int\limits_a^b\vec B \cdot d\vec \ell+ \displaystyle \int\limits_b^c\vec B \cdot d\vec \ell+ \displaystyle \int\limits_c^d\vec B \cdot d\vec \ell+ \displaystyle \int\limits_d^a\vec B \cdot d\vec \ell=\mu_0 (J×L)\)

[ The \(d\vec \ell\) is perpendicular to \(\vec B\) for both path \(b\rightarrow c\) and \(d\rightarrow a\) ]

[ \(\therefore \vec B \cdot d\vec \ell=0\) ]

\(\Rightarrow B\displaystyle \int\limits_a^b\vec d \ell+ 0+ B\displaystyle \int\limits_c^d\ d \ell+ 0 =\mu_0 (J×L)\)

\(\Rightarrow B \cdot L+B \cdot L=\mu_0\;JL\) \(\Big\{ \int\limits_a^b d \ell= \int\limits_c^d d \ell=L \Big\}\)

\(B=\dfrac {\mu_0J}{2}\)

**NOTE:** The magnetic field due to sheet is independent of distance from the sheet.

A \(2\pi×10^{-3}\) \(T\)

B \(8\pi×10^{-3}\) \(T\)

C \(6\pi×10^{-7}\) \(T\)

D \(4\pi×10^{-3}\) \(T\)

- Consider a long hollow cylinder with inner radius "a" and outer radius "b", carrying current \(I\) which is uniformly distributed over the cross sectional area of the conductor.

**Case 1 : **Magnetic field in a region (r < a)

**\(\oint\vec B \cdot d\vec \ell=\mu_0I_{(\text {enclosed)}}\)**

**\(\oint\vec B \cdot d\vec \ell=0\)** [ \(I_{(\text {enclosed)}}=0\), since, the cylinder is hollow ]

**\(B =0\)**

**Case 2 :** Magnetic field in a region (a < r < b)

Current density \(J=\dfrac {I}{\pi(b^2-a^2)}\) \(\Big [J=\dfrac {I}{A} \Big]\)

\(I_{(\text {enclosed)}}=J[\pi (r^2-a^2)]\)

\(I_{(\text {enclosed)}}=\dfrac {I(r^2-a^2)}{(b^2-a^2)}\)

From Ampere's Law,

\(\oint\vec B \cdot d\vec \ell=\mu_0I_{(\text {enclosed)}}\)

\(\Rightarrow B×2\pi\,r=\mu_0×\dfrac {I(r^2-a^2)}{(b^2-a^2)}\)

\(\Big\{\therefore \oint d\ell=2\pi\,r \Big\}\)

\(B=\dfrac {\mu_0\,I\,(r^2-a^2)}{2\pi\,r×(b^2-a^2)}\)

**Case 3 : **Magnetic field in a region (r > b)

From Ampere's Law,

\(\oint\vec B \cdot d\vec \ell=B\cdot 2\pi\,r=\mu_0I\)

\(\Big\{\therefore \oint d\ell=2\pi\,r \Big\}\)

\(B=\dfrac {\mu_0I}{2\pi\,r}\)

A 0, 25 T, 28 T

B 0, 24×10–6 T, 23×10–7 T

C 0, 21.87 ×10–9 T, 33.33 × 10–9 T

D 0, 20 T , 32 ×10–9 T