Magnetic Flux and Time: 5+ Facts You Should Know


In this article, the topic, “magnetic flux and time” with 7 interesting facts will be summarize in a brief manner. Magnetic flux is measurement of amount of magnetism.

Magnetic flux and time both are related to each other.  In a shut in area when magnetic flux is generate for the reason of external magnetic field, the amount magnetic flux is increases while the time is increases. The induced current will be definitely generating a field of the magnetic out of the page.

The direction of the induced current is anti clockwise.

In which point of the compass the bar is move in the reverse point of the compass the induced current will be moved. The magnetic flux also increases with the speed. If a magnet is move faster than the motion of the magnetic flux will be also changes faster, for this particular reason more emf will be produced.

Does magnetic flux depend on time?

Magnetic flux is a vector element and this property is depending upon the direction of the field of the magnetic.  

Yes the property of the magnetic flux is depending on the time. In a closed surface area when the surrounded magnetic field is generate magnetic flux at that time magnetic flux is increases with the rate of the time.

Magnetic flux:-

Magnetic flux is property which defined as, the amount of magnetic field lines which are passes by a given area. It takes measures the measurement of the net amount of magnetic field that passes via a particular specified surface. The area in which magnetic field lines are passes the size does not affect on the property of magnetic flux. In the direction of magnetic field lines magnetic flux is also pointed in that particular direction.

Magnetic flux is denoted by the symbol of, [latex]\phi_B[/latex].

Image – Each point on a surface is associated with a direction, called the surface normal; the magnetic flux through a point is then the component of the magnetic field along this direction;
Image Credit – Wikipedia

Formula of the magnetic flux:-

The formula of the magnetic flux is,

[latex]\phi_B = B.A = BA cos \Theta[\latex]

Derivation of the magnetic flux formula:-

The derivation of the formula is describe below,

[latex]\phi_B[/latex] denoted as magnetic flux

B is denoted as magnetic field

A is denoted as area in which magnetic field lines are passes

[latex]\Theta[/latex] is denoted as the angle in between the magnetic field lines are passes

and the area.

The S.I. unit of the property of magnetic flux is Weber.

The CGS unit of the property of magnetic flux is Maxwell.

The fundamental unit of the property of magnetic flux is Volts – seconds.

 Relation between magnetic flux and time:

The relation with the magnetic flux and time is directly proportional to each other. Means in a shut in area when magnetic flux is generate for the reason of external magnetic field, the amount magnetic flux is increases while the time is increases and the amount of magnetic flux is decreases when the rate of time is decreases.

  • The value of the magnetic flux will be in the lowest point in the case of when the angle of the magnetic field vector and area is almost equal to the value of 90 degree or the value is exactly 90 degree.
  • The value of the magnetic flux will be in the peck point in the case of when the angle of the magnetic field vector and area is almost equal to the value of 0 degree or the value is exactly 0 degree.

How to calculate magnetic flux from time?

With a help of mathematical problem the calculating process of the magnetic flux from the time is discussed below,

Problem: 1:-

The relation expression for the magnetic flux via loop varies is, [latex]\phi = 9t^2 + 8t + C[/latex]. In this expression [latex]\phi[/latex] is present in milliweber C is unchanged and t is present in second.

Now determine the magnitude of induced electromotive force (emf) for the loop at the time period of 3 second.

Solution:-

Given data are, [latex]\phi = 9t^2 + 8t + C[/latex] ………eqn (1)

In this expression \phi is present in milliweber C is unchanged and t is present in second.

 The induced electromotive force for the loop at the time period of 3 second (t = 3) on differentiating the eqn (1) with respect to t.

[latex]\frac{d\phi }{dt} = \frac{d}{dt} 9t^2 + 8t + C[/latex]

[latex]\frac{d\phi }{dt} = 18t + 8[/latex]

At, t = 3

[latex]\frac{d\phi }{dt} = 18 \times (3) + 8[/latex]

[latex]\frac{d\phi }{dt} = 54 + 8[/latex]

[latex]\frac{d\phi }{dt} = 62 mWb/s[/latex]

The relation expression for the magnetic flux via loop varies is, [latex]\phi = 9t^2 + 8t + C[/latex]. In this expression [latex]\phi[/latex] is present in milliweber C is unchanged and t is present in second.

So, the magnitude of induced electromotive force (emf) for the loop at the time period of 3 second is 62 mWb/s.

Problem: 2:-

The relation expression for the magnetic flux via loop varies is, [latex]\phi = 10t^2 + 7t + C[/latex]. In this expression [latex]\phi[/latex] is present in milliweber C is unchanged and t is present in second.

Now determine the magnitude of induced electromotive force (emf) for the loop at the time period of 4 second.

Solution:-

Given data are, [latex]\phi = 10t^2 + 7t + C [/latex]………eqn (1)

In this expression \phi is present in milliweber C is unchanged and t is present in second.

 The induced electromotive force for the loop at the time period of 4 second (t = 3) on differentiating the eqn (1) with respect to t.

[latex]\frac{d\phi }{dt} = \frac{d}{dt} 10t^2 + 7t + C[/latex]

[latex]\frac{d\phi }{dt} = 20t + 7[/latex]

At, t = 4

[latex]\frac{d\phi }{dt} = 20 \times (4) + 7[/latex]

[latex]\frac{d\phi }{dt} = 80 + 7[/latex]

[latex]\frac{d\phi }{dt} = 87 mWb/s[/latex]

The relation expression for the magnetic flux via loop varies is, [latex]\phi = 10t^2 + 7t + C[/latex]. In this expression [latex]\phi[/latex] is present in milliweber C is unchanged and t is present in second.

So, the magnitude of induced electromotive force (emf) for the loop at the time period of 4 second is 87 mWb/s.

Magnetic flux and time graph:

From the graph of magnetic flux and time we easily can observe that respect to time, magnetic flux increases and when the value of time decreases the value of magnetic flux is also decreases, means magnetic flux and time relationship is directly proportional with each other.

magnetic flux and time
Image – Magnetic flux and time graph

Problem statement with solution: 1

A loop which is shaped form like a rectangular. The dimensions of the loop are 0.60 meter and 0.50 meter. The angle in between the magnetic field lines and area is 45 degree and the value of the magnetic field is 0.02 T.

Now determine the value of the magnetic flux for the rectangular loop.

Solution: –

Given data are,

Dimensions of the loop are = 0.60 meter and 0.50 meter.

Magnetic field (B) = 0.02T

The angle in between the magnetic field lines and area [latex](\Theta)[/latex] = 45 degree

We know,

Area (A) = [latex](0.60 \times 0.50)[/latex] meter square = 0.3 meter square

The formula of the magnetic flux is,

[latex]\phi_B = B.A = BA cos \Theta[/latex]

[latex]\phi_B = 0.02 \times 0.3\times cos 45[/latex]

[latex]\phi_B = 0.00312 Wb.[/latex]

A loop which is shaped form like a rectangular. The dimensions of the loop are 0.60 meter and 0.50 meter. The angle in between the magnetic field lines and area is 45 degree and the value of the magnetic field is 0.02 T.

So, the value of the magnetic flux for the rectangular loop is 0.00312 Wb.

Problem statement with solution: – 2

A loop which is a closed circuit resistance carried by 12 ohm. The relation expression for the magnetic flux via loop varies is, [latex]\phi = 10t^2 + 5t + C[/latex]. In this expression \phi is present in milliweber C is unchanged and t is present in second.

Now determine the magnitude of induced electromotive force (emf) for the loop at the time period of 0.15 second.

Solution:-

Given data are,

r = 12 ohm

[latex]\phi = 10t^2 + 5t + C[/latex] ………eqn (1)

In this expression \phi is present in milliweber C is unchanged and t is present in second.

 The induced electromotive force for the loop at the time period of 0.15 second (t = 3) on differentiating the eqn (1) with respect to t.

[latex]e = – \frac{d\phi }{dt} = – \frac{d}{dt} 10t^2 + 5t + C[/latex]

[latex]e = – \frac{d\phi }{dt} = – 20t + 5[/latex]

At, t = 0.15

[latex]e = – \frac{d\phi }{dt} =  – 20 \times (0.15) + 5[/latex]

[latex]\frac{d\phi }{dt} =  2 volt[/latex]

Now,

[latex]i = \frac{e}{R} = \frac{2}{12} = 0.16 A[/latex].

A loop which is a closed circuit resistance carried by 12 ohm. The relation expression for the magnetic flux via loop varies is, [latex]\phi = 10t^2 + 5t + C[/latex]. In this expression [latex]\phi[/latex] is present in milliweber C is unchanged and t is present in second.

So, the magnitude of induced electromotive force (emf) for the loop at the time period of 0.15 second is 0.16 A.

Problem statement with solution: 3

A loop shaped look like a rectangular. The dimensions of the loop are 0.35 meter and 0.49 meter. The angle in between the magnetic field lines and area is 45 degree and the value of the magnetic field is 0.05 T.

Now determine the value of the magnetic flux for the rectangular loop.

Solution: –

Given data are,

Dimensions of the loop are = 0.35 meter and 0.49 meter.

Magnetic field (B) = 0.05T

The angle in between the magnetic field lines and area [latex](\Theta)[/latex] = 45 degree

We know,

Area (A) = [latex](0.35 \times 0.49)[/latex] meter square = 0.1715 meter square

The formula of the magnetic flux is,

[latex]\phi_B = B.A = BA cos \Theta[/latex]

[latex]\phi_B = 0.05 \times 0.1715\times cos 45[/latex]

[latex]\phi_B = 6.06 \times 10^-^0^3 Wb[/latex].

A loop shaped look like a rectangular. The dimensions of the loop are 0.35 meter and 0.49 meter. The angle in between the magnetic field lines and area is 45 degree and the value of the magnetic field is 0.05 T.

So, the value of the magnetic flux for the rectangular loop is [latex]6.06 \times 10^-^0^3 Wb[/latex].

Conclusion:-

The magnetic flux for the reason of external magnetic field via enclosed area increases with time and decreases with time. The relationship between the magnetic flux and time is directly proportional with each other. The induced current must be counterclockwise. If the bar moves in the opposite direction, the direction of the induced current will also be reversed.

Indrani Banerjee

Hi..I am Indrani Banerjee. I completed my bachelor's degree in mechanical engineering. I am a enthusiastic person and I am a person who is positive about every aspect of life. I like to read Books and listening to music.

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