Informative line

Amperes Law

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.

Ampere's Law

  • 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 

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

 

Positive / Negative Direction of Current

  • 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.

Illustration Questions

Which option is incorrect about Ampere's Law from figure shown?

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

×

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.

Hence, option (B) is incorrect.

Which option is incorrect about Ampere's Law from figure shown?

image
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

Option B is Correct

Magnetic Lines

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

Magnetic Field Lines due to Current Carrying Wire

  • 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.

Solenoid

  • 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.

Magnetic Field Lines due to Infinite Sheet

  • 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.

Magnetic Field Lines due to Cylinder

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

Illustration Questions

Choose the correct option for the field lines due to long cylinder.

A

B

C

D

×

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

Hence, option (C) is correct.

Choose the correct option for the field lines due to long cylinder.

A image
B image
C image
D image

Option C is Correct

Amperian Loop

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

Examples

(1) For current carrying wire

  • 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. 

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

 

 

(2) For Ideal Solenoid

  • 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.

Illustration Questions

Choose the correct Amperian loop for long straight wire having steady current \(I\).

A

B

C

D

×

The correct choice of Amperian loop for a current carrying long straight wire should be circular as the magnetic field is also in circular path around wire due to which the direction of magnetic field and \(d\vec \ell\) will be parallel.

Hence, option (A) is correct.

Choose the correct Amperian loop for long straight wire having steady current \(I\).

image
A image
B image
C image
D image

Option A is Correct

Magnetic Field due to Infinite Rod

  • 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.

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

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

\(\Rightarrow\;B×2\pi\,x=\mu_0I\)

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

Illustration Questions

Consider an infinite rod carrying current \(I=2\,A\). Find the magnetic field at P (5, 0) along \(x\)-axis of the rod.

A \(6×10^{–4}\) \((-\hat k)T\)

B \(3 ×10^{–6}\) \(\hat k\;T\)  

C \(0.8 ×10^{–7}\) \((-\hat k)\;T\)  

D \(4 ×10^{–5}\) \(\hat k\;T\)

×

Magnetic field due to an infinite current carrying rod is given as

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

where \(B \)= Magnetic field,

\(I\) = current,

\(x\) = distance of point P from rod

Given : \(I=2\,A\), \(x=5\,m\) 

\(B=\dfrac {4\,\pi×10^{-7}×2}{2\pi\,×5}\)

\(B=0.8×10^{-7}(-\hat k)T\)

Direction of magnetic field at point P is \(-\hat k\) by right hand thumb rule

Consider an infinite rod carrying current \(I=2\,A\). Find the magnetic field at P (5, 0) along \(x\)-axis of the rod.

image
A

\(6×10^{–4}\) \((-\hat k)T\)

.

B

\(3 ×10^{–6}\) \(\hat k\;T\)

 

C

\(0.8 ×10^{–7}\) \((-\hat k)\;T\)

 

D

\(4 ×10^{–5}\) \(\hat k\;T\)

Option C is Correct

Solenoid

  • 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.

Illustration Questions

Calculate the magnetic field in a solenoid of length \(\ell=40\,cm\) having \(N = 400\) turns and current (\(I\)) passing through it is \(5\,A\).

A \(15× 10^{–3}\, T\)

B \(2\pi× 10^{–3}\, T\)

C \(44\ T\)

D \(16× 10^{–3}\, T\)

×

Magnetic field for a solenoid is given as

\(B=\dfrac {\mu_0\,NI}{\ell}\)

where, N = number of turns,

\(I\) = current, 

\(\ell\) = length of solenoid

Given: \( N = 400\)\(\ell =40\,cm\), \(I=5\,A\) 

\(B=\dfrac {4\pi×10^{-7}×400×5}{0.4}\)

\(B =\) \(2\pi× 10^{–3}\, T\) 

Calculate the magnetic field in a solenoid of length \(\ell=40\,cm\) having \(N = 400\) turns and current (\(I\)) passing through it is \(5\,A\).

A

\(15× 10^{–3}\, T\)

.

B

\(2\pi× 10^{–3}\, T\)

C

\(44\ T\)

D

\(16× 10^{–3}\, T\)

Option B is Correct

Magnetic Field at a Point Inside or Outside the Cylinder

  • 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 )

Illustration Questions

Calculate magnetic field at a distance \(r_1=2\,cm\) and \(r_2=6\,cm\) from center of a long cylinder of radius \(R = 5 \,cm\) carrying current \(I=3\,A\) along length.

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

×

For distance r1

Since, \(r_1<R\)

so, magnetic field at a distance r1 is given as 

\(B_1=\left ( \dfrac {\mu_0I}{2\pi\,R^2} \right)r_1\)

image

Given : \( r_1 = 2 \,cm\), \(I=3\,A\), \(R = 5 \,cm\)

\(B_1=\dfrac {4\pi×10^{-7}×3×2×10^{-2}}{2\pi\,×(5×10^{-2})^2}\)

\(B_1=48×10^{-7}\) \(T\)

For distance r2

Since, \(r_2\geq R\)

so, Magnetic field at a distance r2 is given as

\(B_2=\left (\dfrac {\mu_0I}{2\pi\,r_2} \right)\)

image

Given : \(r_2 = 6 \,cm\), \(I=3\,A\)

\(B_2=\dfrac {4\pi×10^{-7}×3}{2\pi\,×6×10^{-2}}\)

\(B_2=10^{-5}\) \(T\)

Calculate magnetic field at a distance \(r_1=2\,cm\) and \(r_2=6\,cm\) from center of a long cylinder of radius \(R = 5 \,cm\) carrying current \(I=3\,A\) along length.

image
A

B1 = 48 × 10 –7 T, B2 = 10 –7

.

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

Option D is Correct

Magnetic Field due to Thick long Finite Sheet

  • 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 (By) due to wires at the left of origin cancel the y component of magnetic field (By) due to wires at the right of origin.
  • Thus, only x-component of magnetic field Bx 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.

Illustration Questions

A thick sheet of current confined in x-z plane with current density J = 3 A /m2. Find the magnetic field above x = 3 m from it.

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 an Amperian loop of rectangular path of length L.

image

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

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

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

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

image

Given : J = 3 A/m2, x = 3 m 

\(B=\dfrac {4\pi×10^{-7}×3}{2}\)

\(B=6\pi×10^{-7}\)\(T\)

image

A thick sheet of current confined in x-z plane with current density J = 3 A /m2. Find the magnetic field above x = 3 m from it.

A

\(2\pi×10^{-3}\) \(T\)

.

B

\(8\pi×10^{-3}\) \(T\)

C

\(6\pi×10^{-7}\) \(T\)

D

\(4\pi×10^{-3}\) \(T\)

Option C is Correct

Magnetic Field for a Long Hollow Cylinder

  • 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,

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

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

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

Illustration Questions

Calculate magnetic field at a distance r = 2 m, 4 m, 6 m , for a hollow cylinder having inner radius a = 3 m and outer radius b = 5 m and current \(I=1\,A\).

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

×

For a region r = 2 m

Since, r < a 

So, by Ampere's Law

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

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

\(B =0\)

image

For a region r = 4 m

Since, a < r < b

The magnetic field in this region is given as 

\(B=\dfrac {\mu_0I(r^2-a^2)}{2\pi\,r(b^2-a^2)}\)

\(B=\dfrac {\mu_0×1×(4^2-3^2)}{2\pi\,×4(5^2-3^2)}\)

\(B=\dfrac {4\pi×10^{-7}×7}{2\pi\,×4×16}\)

\(B=\dfrac {7×10^{-7}}{32}\\=21.87×10^{-9} \,T\)

image

For a region r = 6 m

Since, r > b,

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

\(B=\dfrac {4\pi×10^{-7}×1}{2\pi×6}\)

\(B=\dfrac {10^{-7}}{3}\)

\(B= 33.33×10^{-9}\) \(T\)

image

Calculate magnetic field at a distance r = 2 m, 4 m, 6 m , for a hollow cylinder having inner radius a = 3 m and outer radius b = 5 m and current \(I=1\,A\).

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

Option C is Correct

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