
A coil having $1000$ number of turns and area $0.02{m^2}$ is placed perpendicular to a magnetic field $5 \times {10^{ - 3}}T$. Calculate flux passing through this coil.
Answer
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Hint: We can calculate the magnetic flux that is passing through the coil by using the formula $\phi = NBA\cos \theta $. Here the values of the various parameters are given in the question and $\theta $ is the angle between the area vector and the magnetic field which is $0^\circ $
Formula Used In this solution we will be using the following formula,
$\phi = NBA\cos \theta $
where $\phi $ is the magnetic flux through the coil
$N$ is the number of turn
$B$ is the magnitude of the magnetic field
$A$ is the area of the coil and
$\theta $ is the angle between the area vector and the magnetic field
Complete Step by Step Solution:
The magnitude of the flux that is passing through a coil that is passed in a magnetic field is given by the formula,
$\phi = NBA\cos \theta $
In this problem we are told that the coil is placed perpendicular to the flow of the magnetic field.
So the area vector of any surface is in the direction of the perpendicular to the plane that the surface lies on. Therefore, since the surface of the coil is perpendicular to the magnetic field, so the area vector of the surface of the coil will be parallel to the magnetic field. That is the angle between the area vector and the magnetic field is $\theta = 0^\circ $.
Now the value of the magnetic field is given in the question as, $B = 5 \times {10^{ - 3}}T$ and the surface area of the coil is given as, $A = 0.02{m^2}$. The number of turns of the coil is, $N = 1000$
So by substituting these values we get the flux passing through this coil as,
$\phi = 1000 \times 5 \times {10^{ - 3}} \times 0.02\cos 0^\circ $
The value of $\cos 0^\circ $ is 1. So, we get
$\phi = 5 \times 0.02 \times 1$
Therefore, the flux passing through this coil is, $\phi = 0.1Wb$.
Note: The magnetic flux passing through a coil is the number of the magnetic field lines that is passing through the closed surface of the coil. Here the changing area under consideration will change the magnetic field lines that are passing through the coil.
Formula Used In this solution we will be using the following formula,
$\phi = NBA\cos \theta $
where $\phi $ is the magnetic flux through the coil
$N$ is the number of turn
$B$ is the magnitude of the magnetic field
$A$ is the area of the coil and
$\theta $ is the angle between the area vector and the magnetic field
Complete Step by Step Solution:
The magnitude of the flux that is passing through a coil that is passed in a magnetic field is given by the formula,
$\phi = NBA\cos \theta $
In this problem we are told that the coil is placed perpendicular to the flow of the magnetic field.
So the area vector of any surface is in the direction of the perpendicular to the plane that the surface lies on. Therefore, since the surface of the coil is perpendicular to the magnetic field, so the area vector of the surface of the coil will be parallel to the magnetic field. That is the angle between the area vector and the magnetic field is $\theta = 0^\circ $.
Now the value of the magnetic field is given in the question as, $B = 5 \times {10^{ - 3}}T$ and the surface area of the coil is given as, $A = 0.02{m^2}$. The number of turns of the coil is, $N = 1000$
So by substituting these values we get the flux passing through this coil as,
$\phi = 1000 \times 5 \times {10^{ - 3}} \times 0.02\cos 0^\circ $
The value of $\cos 0^\circ $ is 1. So, we get
$\phi = 5 \times 0.02 \times 1$
Therefore, the flux passing through this coil is, $\phi = 0.1Wb$.
Note: The magnetic flux passing through a coil is the number of the magnetic field lines that is passing through the closed surface of the coil. Here the changing area under consideration will change the magnetic field lines that are passing through the coil.
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