Informative line

Change In Magnetic Flux

Practice flux remains constant through change in magnetic flux due to a current carrying wire, and calculate the total magnetic flux through the loop due to current in the wire and through a coil.

Magnetic Flux

  • Magnetic Flux

          Magnetic flux is the amount of magnetic field passing through a surface.

         In other words, it is the total number of magnetic lines of field passing through a specified area in magnetic field.

  • Symbol of Magnetic Flux 

             It  is denoted by \('\phi _m'\).

  •   Magnetic Field depends upon

           (i)  Area:  As the area increases, the number of field lines passing through loop increases.

           Thus , \(\phi _m \propto\, A\)   

(ii) Magnetic Field: 

  As  B increases , the number of field lines passing through loop increases .

  Thus ,  \(\phi _m \propto\, B\) 

(iii)  Angle between Area and Magnetic Field:

       As the angle between area and magnetic field increases, the number of field lines passing through loop increases. 

      Thus, \(\phi _m \propto\, \) Angle between area and magnetic field

Illustration Questions

Change in magnetic flux with time does not depend upon

A Change in area, constant B, constant alignment

B Change in B, constant area, constant alignment

C Change in alignment, constant B, constant area

D Non - uniform but constant B, constant area, constant alignment

×

Magnetic field, area and angle between magnetic field and area are constant .

i.e., B, A, \(\theta\) are constant .

\(\Rightarrow\phi _m\) is constant .

\(\Rightarrow\dfrac{d\,\phi_m}{dt} = 0\)

Hence , option (D) is correct.

Change in magnetic flux with time does not depend upon

A

Change in area, constant B, constant alignment

.

B

Change in B, constant area, constant alignment

C

Change in alignment, constant B, constant area

D

Non - uniform but constant B, constant area, constant alignment

Option D is Correct

Magnetic Flux in Presence of Magnet

  • Consider a conducting coil by which galvanometer is connected.
  • A bar magnet is placed, as shown in figure.

 

Different cases can be applied

  • Case 1 : When bar magnet is stationary.

          The magnetic flux linked with coil will be constant as the number of field lines passing through coil is constant.

Case 2 : When bar magnet is moved closer to coil.

The magnetic flux linked with coil will increase as the number of field lines passing through coil is increasing.

Case 3: When bar magnet is moved away from coil.

The magnetic flux linked with coil will decrease as the number of field lines passing through coil is decreasing.

Illustration Questions

In which of the following cases, the flux does not change through the coil?

A

B

C

D

×

Option (A) is incorrect because the relative velocity of coil and bar magnet is not zero.

Thus, flux changes through the coil.

image

Option (C) is incorrect because the relative velocity of coil and bar magnet is not zero.

Thus, flux changes through the coil.

image

Option (D) is incorrect because the relative velocity of coil and bar magnet is not zero.

Thus, flux changes through the coil.

image

Option (B) is correct because the relative velocity of coil and bar magnet is zero .

Thus, flux does not change through the coil.

image

In which of the following cases, the flux does not change through the coil?

A image
B image
C image
D image

Option B is Correct

Change in Magnetic Flux due to a Current Carrying Wire

  • Change in Magnetic Flux depends upon

            (i) Change in field with respect to time.

            (ii) Change in orientation and area with respect to time of any loop or surface.

  • Consider a current element  AB of straight wire carrying current \(I\).
  • The flux is to be analyzed over a square loop, as shown in figure.

 

  • The magnetic field is non-uniform but constant with respect to time.

            Thus, magnetic flux is constant.

  • Change in Flux will take place in following situation :

    (i) When loop and wire are not stationary with respect to each other.

    (ii) Current in the wire is the function of time i.e., increases or decreases with respect to time.  

  • In case of solenoid:

         The change in flux through loop depends upon the current in the solenoid.

Illustration Questions

In which of the following cases, the flux remains constant?

A

B

C

D

×

Option (A) is incorrect because current is a function of time. Hence, flux is not constant. 

image

Option (B) is incorrect because current is a function of time . Hence, flux is not constant. 

image

Option (C) is incorrect because loop is not stationary with respect to wire. 

image

Option (D) is correct because current is constant and loop and solenoid are stationary to each other. 

image

In which of the following cases, the flux remains constant?

A image
B image
C image
D image

Option D is Correct

Illustration Questions

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 \(\phi _m = {\mu_0 \,I a}\, \ell n \left (1+\dfrac{a}{d}\right ) \)

B \(\phi _m =\dfrac {\mu_0 \,I a}{2\pi} \ell n \left (1+\dfrac{a}{d}\right ) \)

C \(\phi _m = {\mu_0}\,{2\pi}\, \ell n \,(I\,a) \)

D \(\phi _m =\dfrac {\mu_0 \,2\pi}{a} \ell n \left (1+\dfrac{a}{d}\right ) \)

×

Magnetic flux through a current carrying loop is given as 

\(\phi _m = \displaystyle \int \vec B \cdot d \vec A \)

\(\phi _m = \displaystyle \int B \,d A \)                   {cos 0º = 1 }

\(\phi _m = \displaystyle \int \dfrac{\mu_0 I }{2\pi x} dA\)                   \(\left [\therefore B = \dfrac{\mu_0I}{2\pi x}\right]\)  

image

A small element of area dA  

{ dA=adx }

\(\phi_m = \dfrac{\mu_0 \,I}{2\pi} \displaystyle \int \limits _d^{d+a} \dfrac{a\,dx}{x}\)

Due to current carrying wire, magnetic field will be function \(dx\).

\(\phi_m = \dfrac{\mu_0 I a}{2\pi} \displaystyle \int \limits _d^{d+a} \dfrac{1}{x}\,dx\)

\(\phi_m = \dfrac{\mu_0 I a}{2\pi} \displaystyle \big[\ell n \,x\big]_d^{d+a}\)

\(\phi_m = \dfrac{\mu_0 Ia}{2\pi} \ell n \left(1 + \dfrac{a}{d}\right) \)

image

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.

image
A

\(\phi _m = {\mu_0 \,I a}\, \ell n \left (1+\dfrac{a}{d}\right ) \)

.

B

\(\phi _m =\dfrac {\mu_0 \,I a}{2\pi} \ell n \left (1+\dfrac{a}{d}\right ) \)

C

\(\phi _m = {\mu_0}\,{2\pi}\, \ell n \,(I\,a) \)

D

\(\phi _m =\dfrac {\mu_0 \,2\pi}{a} \ell n \left (1+\dfrac{a}{d}\right ) \)

Option B is Correct

Effect of Flipping of Coil in Magnetic Field on Magnetic Flux 

  • Consider a coil of area A which is placed in uniform magnetic field.
  • The coil is flipped by some angle. 

Effect of Flipping

   The magnetic field and area remains constant but due to flipping, the magnetic flux through the coil changes as the angle between the area vector and magnetic field changes.

  •  At initial position, magnetic flux is

             \((\phi_m)_i = B\,A \,cos0º\)

            \((\phi_m)_i = B\,A \)

    

  • Let it be flipped by 180º.
  • At final position, magnetic flux is

           \((\phi_m)_f = B\,A \,cos180º\)

           \((\phi_m)_f = -B\,A \)

     

  • Change in flux

          \(\Delta \phi_m = |(\phi_m)_f -(\phi_m)_i|\)

          \(\Delta \phi_m = | -BA -BA |\)

          \(\Delta \phi_m = 2\,BA \)

Illustration Questions

A circular loop of radius r = 1 m is placed in uniform magnetic field B = 1 T, as shown in figure . If the circular loop is flipped by 180º, then find the total change in flux.

A \(3\, \pi \,Wb\)

B \(2\, \pi \,Wb\)

C \(5\, \pi \,Wb\)

D \(0\)

×

Radius of circular loop, r = 1 m 

Angle by which loop is flipped, \(\theta = \) 180º 

Magnetic field, \(\vec B = 1\,T\)

Magnetic flux before flipping

\((\phi_m)_i = BA\)

Magnetic flux after flipping 

  \((\phi_m)_f = -BA\)

Change in flux,

 \(\Delta \phi_m = |(\phi_m)_f - (\phi_m)_i | \)

 \(\Delta \phi_m = |-BA - BA | \)

 \(\Delta \phi_m = 2\,BA \)

Change in flux, \(\Delta \phi_m = 2\,BA\)

                                    \(= 2(1) \,[\pi(1)^2]\)

                                    \(= 2\pi \, \;Wb\)

A circular loop of radius r = 1 m is placed in uniform magnetic field B = 1 T, as shown in figure . If the circular loop is flipped by 180º, then find the total change in flux.

image
A

\(3\, \pi \,Wb\)

.

B

\(2\, \pi \,Wb\)

C

\(5\, \pi \,Wb\)

D

\(0\)

Option B is Correct

Magnetic Flux when Magnetic Field is a Function of Time 

  • Consider a square loop of side 'a'  and the magnetic field is given as a function of time.

              \(\vec B = B_0\,t\) 

  • The magnetic flux through the square loop is given by 

           \(\phi_m = B . (Area \,of \,loop)\)

           \(\phi_m = B_0 \,t \,(a^2) \,cos\,\theta\)

where,

\(\theta\) is the angle between area vector and magnetic field vector ,as shown in figure.

\(\phi_m = (B_0 \,t) a^2 \,cos\,\theta\)

 

Illustration Questions

If magnetic field is \(\vec B = B_0 \,sin \,\omega t\), then what will be the flux linked with a square loop of side \(\ell \) at time t , when loop is placed perpendicular to the magnetic field?

A \(\ell ^2 \, B _0 \, sin \,\omega t\)

B \(\dfrac{\ell ^2}{B _0} \, sin \,\omega t\)

C \(\dfrac{\ell ^2\, B _0}{sin \,\omega t} \, \)

D \(\dfrac{ B _0 sin \,\omega \,t \, \ell _0\,}{2} \)

×

Length of side of square loop = \(\ell\)

Magnetic field, \(\vec B = B_0 \,sin \,\omega t \,\hat n\) 

Area vector of square loop, \(\vec A = \ell^2 \, \hat n\) 

time = t

Magnetic flux is given by,

\(\phi_m = \vec B . \vec A\)

\(\phi _m = B_0 \, \ell^2 \, sin\, \omega t \,cos 0º \)

\(\phi _m = B_0 \, \ell^2 \, sin\, \omega t \)

If magnetic field is \(\vec B = B_0 \,sin \,\omega t\), then what will be the flux linked with a square loop of side \(\ell \) at time t , when loop is placed perpendicular to the magnetic field?

image
A

\(\ell ^2 \, B _0 \, sin \,\omega t\)

.

B

\(\dfrac{\ell ^2}{B _0} \, sin \,\omega t\)

C

\(\dfrac{\ell ^2\, B _0}{sin \,\omega t} \, \)

D

\(\dfrac{ B _0 sin \,\omega \,t \, \ell _0\,}{2} \)

Option A is Correct

Magnetic Flux of Coil Placed in Solenoid

  • Consider a solenoid is connected to AC source.
  • The current  in turns of solenoid is given by 

             \(I = I _0 \,sin\,\omega t\)

  • Magnetic field through solenoid 

            \(B=\mu _0 \, n \,I\)

           \(B = \mu_0 \, n \, I_0\, sin\,\omega t\)

where,

        n is the number of turns per unit length .

  • Consider a coil of area A, placed in the solenoid as shown in figure.

  • Magnetic Flux through the Coil

          \(\phi_m = B.A\)

          \(\phi_m = BA \, cos\,0º\)

          \(\phi_m = A \, \mu_0\, n \,I_0 \, sin \,\omega t\)

 

Illustration Questions

A solenoid of diameter d1 = 4 cm is wound with n = 1000 turns per meter. The current through the solenoid is \(I = 3 \,sin\,\omega t\). If a loop of diameter d2 = 2 cm is placed at the center of solenoid, find the flux through loop. [\(\mu_0 = 4\pi × 10^{-7} N/A^2\)]

A \(0.12\, \pi^2 \, sin\,\omega t\, \mu Wb\)

B \(0.15\, \pi^2 \, sin\,\omega t\, \mu Wb\)

C \(0.30\, \pi^2 \, sin\,\omega t\, \mu Wb\)

D \(\pi^2 \, sin\,\omega t\, \mu Wb\)

×

Diameter of solenoid, d1 = 4 cm

Number of turns per meter, n =1000

Current through solenoid, \(I= 3 \,sin\,\omega t\) 

Diameter of loop, d2 = 2 cm

Magnetic field through solenoid, \(B = \mu_0\, n\,I\)

                                                  = \(4\pi × 10^{-7} × 1000 × 3 \,sin \,\omega t\)

Area of loop,  \(A = \pi (1)^2 × 10^{-4}\, m\)

Magnetic flux, \(\phi_m = B.A\)

\(\phi_m = 4\pi × 10^{-7} × 1000 × 3 \,sin\,\omega t × \pi× 10^{-4}\)

\(\phi_m = 0.12\, \pi^2 \, sin\,\omega t\, \mu Wb\)

 

A solenoid of diameter d1 = 4 cm is wound with n = 1000 turns per meter. The current through the solenoid is \(I = 3 \,sin\,\omega t\). If a loop of diameter d2 = 2 cm is placed at the center of solenoid, find the flux through loop. [\(\mu_0 = 4\pi × 10^{-7} N/A^2\)]

A

\(0.12\, \pi^2 \, sin\,\omega t\, \mu Wb\)

.

B

\(0.15\, \pi^2 \, sin\,\omega t\, \mu Wb\)

C

\(0.30\, \pi^2 \, sin\,\omega t\, \mu Wb\)

D

\(\pi^2 \, sin\,\omega t\, \mu Wb\)

Option A is Correct

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