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.

#### 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?

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.

#### 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
B
C
D

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.

#### 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 )

#### 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

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

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

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

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.

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

# 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.

#### 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 $$I$$.

A
B
C
D

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.

$$\oint B\cdot dx=\mu_0I$$

$$\Rightarrow\;B\oint dx=\mu_0I$$

$$\Rightarrow\;B×2\pi\,x=\mu_0I$$

$$B=\dfrac {\mu_0I}{2\pi\,x}$$

#### 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.

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

# 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.

#### 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.

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

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

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

$$B=\dfrac {\mu_0J}{2}$$

Given : J = 3 A/m2, x = 3 m

$$B=\dfrac {4\pi×10^{-7}×3}{2}$$

$$B=6\pi×10^{-7}$$$$T$$

### 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,

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

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

$$B=\dfrac {\mu_0I}{2\pi\,r}$$

#### 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) ]

$$B =0$$

For a region r = 4 m

Since, a < r < b

The magnetic field in this region is given as

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

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

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

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

For a region r = 6 m

Since, r > b,

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

$$B=\dfrac {4\pi×10^{-7}×1}{2\pi×6}$$

$$B=\dfrac {10^{-7}}{3}$$

$$B= 33.33×10^{-9}$$ $$T$$

### 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