Informative line

### Magnetic Flux

Learn magnetic flux definition with equation, practice to calculate the total magnetic flux through the loop due to current in the wire and find electric flux through cylinder and magnetic field of a current loop.

# Concept of Flux

• Flux of any physical quantity is the measure of flow of that quantity passing through any specific area.
• To understand the idea of flux, consider a region in which rain is falling vertically downwards and ring is placed in this region.
• The rate of flow of rain passing through area of the ring depends on its area and orientation of ring.

## Case I : "Rings of different area and same orientation"

Since,

Area c > Area b > Area a .

Hence, flux through

c > b > a

## Case II : "Rings of same area but different orientation"

• Plane of ring (a) oriented perpendicular to field.
• Plane of ring (b) oriented at an angle to the normal of the field.
• Plane of ring (c) oriented along the field.
• Flux through Ring (a) > Ring (b) > Ring (c).
• Flux of Ring (c) is zero as its plane is oriented along the field so, no rain is passing through its area.

## Magnetic Flux

Magnetic flux is defined as total number of magnetic lines passing through a specified area placed in a magnetic field.

### SI Unit of Magnetic Flux

The SI unit of magnetic flux is Weber (Wb) or T.m2 (Tesla.m2).

#### Three identical rings of area A are placed in an uniform magnetic field, as shown in figure. Choose the correct option regarding to flux linked with them.

A a > b > c

B a = b = c

C c > b > a

D a > c > b

×

Area of rings is same, but orientation is different.

Number of field lines passing through A > Number of field lines passing through B > Number of field lines passing through C

Hence, flux linkage is in order :

Ring (a) > Ring (b) > Ring (c)

### Three identical rings of area A are placed in an uniform magnetic field, as shown in figure. Choose the correct option regarding to flux linked with them.

A

a > b > c

.

B

a = b = c

C

c > b > a

D

a > c > b

Option A is Correct

# Flux linked with an Area

• Consider a closed surface of cylinder placed in a non-uniform magnetic field as shown in figure.
• Area of P = Area of Q.
• Number of field lines associated with P is greater than number of field lines associated with Q.
• Flux linked with P is greater than flux linked with Q.
• Flux is denoted by $$\phi$$.

$$\phi_P>\phi_Q$$

## Entering and leaving concept of field lines for calculation of flux

• Consider a body placed in a magnetic field.
• Calculate number of field lines entering and leaving the body.

Case 1: If the number of field lines entering and leaving the body are equal, then total flux linked with the body will be zero.

Case 2:  If the number of field lines entering and leaving the body are not equal, then

Net flux = Number of field lines leaving - Number of field lines entering

## Concept of projected area

• When any body is placed in an uniform magnetic field such that its surface is perpendicular to the magnetic field, then magnetic flux associated with it, is maximum.
• Now if, surface of same body is placed at some angle with the magnetic field rather than perpendicular, then magnetic flux associated with it, will be less than maximum value.
• The number of magnetic field lines passing through the tilted surface is less than the number of field lines passing through perpendicular surface but same as the number of field lines passing through the projection of area of surface.

Number of field lines passing through tilted surface (A) = Number of field lines passing through projection of area of surface

$$\phi_{m'_\bot}=\phi_{m'}<\phi_m$$

$$A_{m'_\bot} < A_{m'} = A_m$$

#### Choose the correct option regarding flux linkage among surface A and surface B of the cylinder placed in a non-uniform magnetic field, as shown in figure.

A $$\phi_A>\phi_B$$

B $$\phi_B>\phi_A$$

C $$\phi_A=\phi_B$$

D

×

Number of field lines passing through (A)  > Number of field lines passing through (B)

Hence, flux linkage $$\phi_A>\phi_B$$

### Choose the correct option regarding flux linkage among surface A and surface B of the cylinder placed in a non-uniform magnetic field, as shown in figure.

A

$$\phi_A>\phi_B$$

.

B

$$\phi_B>\phi_A$$

C

$$\phi_A=\phi_B$$

D

Option A is Correct

# Mathematical Form of Flux

## Area Vector

• By convention, area vector is always taken outwards / perpendicular to the plane, as shown in figure.
• Consider a closed surface, as shown in figure. Area vector for all the three surfaces A, B and C of cylinder is taken outwards/perpendicular to the surface.

## Flux

Consider a ring placed in a uniform magnetic field, as shown in figure.

• Magnetic flux linked with the area A is given as the dot product of magnetic field and area vector.
• Flux linked with area A

$$\phi_B=B(A\;cos\theta)$$

$$\phi_B=\vec B\cdot\vec A$$

• For non-uniform magnetic field, magnetic field may vary over a large surface.
• The concept of flux is meaningful for small area where magnetic field is approximately constant, when total surface is placed in non-uniform field.

$$\Delta\phi_B=B\cdot(\Delta A\;cos\theta)$$

or, $$\Delta\phi_B=\vec B\cdot\Delta \vec A$$

where $$\Delta\phi_B$$ is the magnetic field through this element

• The magnetic flux through whole surface is given as summation of magnetic flux through all these small elements.

$$\phi_B=\sum\vec B\cdot\Delta\vec A$$

• If area of each element approaches zero, the number of elements approach infinity, then the sum is replaced by integral.

$$\phi_B=\int_s\vec B\cdot d\vec A$$

• For closed surface,  $$\phi_B=\int\vec B\cdot d\vec A$$
• If in a region magnetic field is given as  $$\vec B=B_x\hat i+B_y\hat j+B_z\hat k$$
• An area A is placed in an magnetic field with area vector $$\vec A=A_x\hat i+A_y\hat j+A_z\hat k$$
• Flux linked with this area is given as $$\phi=\vec B\cdot\vec A$$

$$\phi=B_xA_x+B_yA_y+B_zA_z$$

#### Calculate the amount of flux linked with area vector $$\vec A = (3\hat i+4\hat j)$$ m2  which is placed in an magnetic field $$\vec B = (4\hat i+6\hat j)$$ T.

A 0 Wb

B 36 Wb

C 30 Wb

D 400 Wb

×

Flux linked with area A is given as  $$\phi=\vec B\cdot\vec A$$

Given: $$\vec B = (4\hat i+6\hat j)$$ T, $$\vec A = (3\hat i+4\hat j)$$ m2

$$\phi=(4\hat i+6\hat j)\cdot(3\hat i+4\hat j)$$

$$\phi=$$12 + 24

= 36 Wb

### Calculate the amount of flux linked with area vector $$\vec A = (3\hat i+4\hat j)$$ m2  which is placed in an magnetic field $$\vec B = (4\hat i+6\hat j)$$ T.

A

0 Wb

.

B

36 Wb

C

30 Wb

D

400 Wb

Option B is Correct

# Calculation of Flux when Angle is Given

## Area Vector

• By convention, area vector is always taken outwards / perpendicular to the plane, as shown in figure.
• Consider a closed surface, as shown in figure. Area vector for all the three surface A, B and C of cylinder is taken outwards/perpendicular to the surface.

## Flux

Consider a ring placed in a uniform magnetic field, as shown in figure.

• Magnetic flux linked with the area A is given as the dot product of magnetic field and area vector.
• Flux linked with area A

$$\phi_B=B(A\;cos\theta)$$

$$\phi_B=\vec B\cdot\vec A$$

#### Calculate magnetic flux $$(\phi_ m)$$ through surface given $$\vec B=40\,T$$, $$\theta=37°$$ and area $$A = 15 ×10\;m^2$$, as shown in figure, if $$\vec B$$ is uniform. [ $$cos \;37°=\dfrac{4}{5}$$ ]

A 2400 Wb

B 4800 Wb

C 1600 Wb

D 1800 Wb

×

Magnetic flux $$(\phi_ m)$$ through a given surface (A) placed in a magnetic field (B) is given as,

$$\phi _m=B(A\;cos\theta)$$

Given:

$$B = 40\, T$$$$\theta=37°$$$$A = 15 ×10 \;m^2$$

$$\phi_ m=40\times15\times10\times cos37°$$

$$\phi_ m=40\times15\times10\times\dfrac {4}{5}$$      $$\left[cos37°=\dfrac{4}{5}\right]$$

$$\phi_ m=4800 \,Wb$$

### Calculate magnetic flux $$(\phi_ m)$$ through surface given $$\vec B=40\,T$$, $$\theta=37°$$ and area $$A = 15 ×10\;m^2$$, as shown in figure, if $$\vec B$$ is uniform. [ $$cos \;37°=\dfrac{4}{5}$$ ]

A

2400 Wb

.

B

4800 Wb

C

1600 Wb

D

1800 Wb

Option B is Correct

# Flux Linked with Cylindrical Body

• Consider a cylindrical body placed in a uniform magnetic field $$\vec B$$.
• To calculate flux, by convention area vector is always taken perpendicular and in outward direction to the surface.

• Area vector of three surfaces of a cylindrical body, are shown in figure.

### Flux through A1

• By convention, area vector is always taken perpendicular and in outward direction to the surface.
• Angle between $$\vec B$$ and $$d\vec A_1$$ is 180°.

$$\phi_1=\vec B\cdot d\vec A_1$$

$$\phi_1=B\;dA_1\;cos(180°)$$

$$\phi_1=-B\ dA_1$$

• It can be concluded that for surface 1 by convention, we have taken the $$d\vec A$$ outside the surface which is in opposite direction to magnetic field.
• Hence, the flux coming out of surface 1 is negative.

Conclusion: For any particular surface through which the magnetic field lines are entering, flux will be negative.

### Flux through A2

Angle between $$\vec B$$ and $$d\vec A_2$$ is 90°.

$$\phi_2=\vec B\cdot d\vec A_2$$

or, $$\phi_2=B\;dA_2\;cos(90°)$$

or, $$\phi_2=0$$

Conclusion: If magnetic field and area vector are perpendicular to each other, then flux through that surface is zero.

### Flux through A3

Angle between $$\vec B$$ and $$d\vec A$$ is 0°.

$$\phi_1=\vec B\cdot d\vec A$$

$$\phi_1=B\;dA\;cos0°$$

$$\phi_1=B\cdot dA$$

• By convention, for surface 3 we have taken $$d\vec A$$ outside the surface which is in the direction of magnetic field.
• Hence, flux coming out of surface 3 is positive.

Conclusion: For any particular surface through which the magnetic field lines are leaving, flux will be positive.

#### A cube is placed in a uniform magnetic field $$\vec B=B_x\hat i+B_z\hat k$$ T. Determine the sign of flux through ABCD, CDHG and EFGH.

A $$B_zA_1,\,-B_xA_2,\,+B_zA_3$$

B $$B_zA_1,\,B_xA_2,\,-B_zA_3$$

C $$B_zA_1,\,-B_xA_2,\,-B_zA_3$$

D $$-B_zA_1,\,B_xA_2,\,B_zA_3$$

×

Magnetic field is present in +X and +Z direction.

For face ABCD

Direction of area vector

$$\hat n_1=\hat k$$

$$\vec A_1=A_1\;\hat k\;m^2$$

$$\vec B=(B_x\hat i + B_z\;\hat k)$$ T

Total flux  $$\phi_1=\vec B\cdot\vec A_1$$

$$\phi_1=(B_x\hat i +B_z \hat k)\cdot(A_1\hat k)$$

$$\phi_1=B_zA_1=$$ positive

For face CDHG

Direction of area vector

$$\hat n_2=\hat i$$

$$\vec A_2=(A_2\;\hat i)\;m^2$$

$$\vec B=(B_x\hat i + B_z\;\hat k)$$ T

Total flux  $$\phi_2=\vec B\cdot\vec A_2$$

$$\phi_2=(B_x\hat i +B_z \hat k)\cdot(A_2\hat i)$$

$$\phi_2=B_xA_2=$$ positive

For face EFGH

Direction of area vector

$$\hat n_3=-\hat k$$

$$\vec A_3=(-A_3\;\hat k)\;m^2$$

$$\vec B=(B_x\;\hat i + B_z\;\hat k)$$ T

Total flux  $$\phi_3=\vec B\cdot\vec A_3$$

$$\phi_3=(B_x\hat i +B_z \hat k)\cdot(-A_3\hat k)$$

$$\phi_3=-B_z\;A_3=$$ negative

### A cube is placed in a uniform magnetic field $$\vec B=B_x\hat i+B_z\hat k$$ T. Determine the sign of flux through ABCD, CDHG and EFGH.

A

$$B_zA_1,\,-B_xA_2,\,+B_zA_3$$

.

B

$$B_zA_1,\,B_xA_2,\,-B_zA_3$$

C

$$B_zA_1,\,-B_xA_2,\,-B_zA_3$$

D

$$-B_zA_1,\,B_xA_2,\,B_zA_3$$

Option B is Correct

# Magnetic Flux in a Current Carrying Loop

• The magnetic field due to current carrying wire is non-uniform.
• The magnetic field at a point near the wire is more.

## Magnetic Flux

Magnetic flux through a current carrying loop is given as

$$\phi _m=\int\vec B.d\vec A$$

where  $$\vec B$$ = magnetic field

$$d\vec A$$ = Area vector

$$\phi _m$$ = magnetic flux

#### A square loop of side 'a' is located near a long wire carrying current $$I$$. The distance between wire and closest side of loop is d such that they are parallel to each other. Calculate the total magnetic flux through the loop due to current in the wire.

A $$\mu_0Ia\,\ell n\left(1+\dfrac{a}{d}\right)$$

B $$\dfrac{\mu_0Ia}{2\pi}\,\ell n\left(1+\dfrac{a}{d}\right)$$

C $$\mu_02\pi\,\ell n(Ia)$$

D $$\dfrac{\mu_02\pi}{a}\,\ell n\left(1+\dfrac{a}{d}\right)$$

×

Magnetic flux through a current carrying loop is given as

$$\phi _m=\int\vec B.d\vec A$$

$$\phi _m=\int\ B\,d A$$

$$\displaystyle\phi _m=\int\dfrac{\mu_0I}{2\pi x}d A$$      $$\left[\therefore \,B=\dfrac{\mu_0I}{2\pi x}\right]$$

A small element of area dA  is taken as shown in figure.    $$\{dA=adx\}$$

$$\displaystyle\phi _m=\dfrac{\mu_0I}{2\pi} \int\limits^{d+a}_d \dfrac{a\,d x}{x}$$

Due to current carrying wire, magnetic field will be function of x

$$\displaystyle\phi _m=\dfrac{\mu_0Ia}{2\pi} \int\limits^{d+a}_d \dfrac{1}{x}dx$$

$$\displaystyle\phi _m=\dfrac{\mu_0Ia}{2\pi} \left[\ell nx\right]^{d+a}_d$$

$$\displaystyle\phi _m=\dfrac{\mu_0Ia}{2\pi} \ell n\left(1+\dfrac{a}{d}\right)$$

### A square loop of side 'a' is located near a long wire carrying current $$I$$. The distance between wire and closest side of loop is d such that they are parallel to each other. Calculate the total magnetic flux through the loop due to current in the wire.

A

$$\mu_0Ia\,\ell n\left(1+\dfrac{a}{d}\right)$$

.

B

$$\dfrac{\mu_0Ia}{2\pi}\,\ell n\left(1+\dfrac{a}{d}\right)$$

C

$$\mu_02\pi\,\ell n(Ia)$$

D

$$\dfrac{\mu_02\pi}{a}\,\ell n\left(1+\dfrac{a}{d}\right)$$

Option B is Correct