Question

A coil of \[100\] turns and \[5\] square centimetres is placed in a magnetic field \[B = 0.2\,{\text{T}}\] . The normal to the plane of the coil makes an angle of \[60^\circ \] with the direction of the magnetic field. The magnetic flux linked with the coil is:
A. \[5 \times {10^{ - 3}}\,{\text{Wb}}\]
B. \[5 \times {10^{ - 5}}\,{\text{Wb}}\]
C. \[{10^{ - 2}}\,{\text{Wb}}\]
D. \[{10^{ - 4}}\,{\text{Wb}}\]

Answer 1

Hint: First of all, we will convert the area unit into S.I units, then we will simply use the formula which states that magnetic flux is directly proportional to the magnetic field, area of cross section of the wire and number of turns. We will simply substitute the values and manipulate accordingly to obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
Total number of turns in the coil is \[N = 100\] .
The area of the cross section of the coil which is used for the windings is \[A = 5\,{\text{c}}{{\text{m}}^2}\] .
The magnetic field produced by the coil is \[B = 0.2{\text{T}}\] .
We are asked to find the magnetic field linked with the coil.

We convert the unit of area into S.I unit first:
\[
A = 5\,{\text{c}}{{\text{m}}^2} \\
\Rightarrow A = 5 \times {\left( {{{10}^{ - 2}}} \right)^2}\,{{\text{m}}^2} \\
\Rightarrow A = 5 \times {10^{ - 4}}\,{{\text{m}}^2} \\
\]
To begin with, according to the question, the normal to the plane of the coil makes an angle of \[60^\circ \] with the direction of the magnetic field which is produced in the coil.
So, to find the magnetic flux, we apply the formula which is given below:
\[\phi = N \times B \times A \times \cos \theta \] …… (1)
Where,
\[N\] indicates the number of turns in the coil.
\[B\] indicates the magnetic field produced in the coil.
\[A\] indicates the area of the cross of the coil which is used for the windings.
\[\theta \] indicates the angle at which the normal to the plane makes with the direction of the magnetic field which is produced in the coil.

Now, we substitute the required values in the equation (1) and we get:
\[
\Rightarrow \phi = 100 \times 0.2 \times 5 \times {10^{ - 4}} \times \cos 60^\circ \\
\Rightarrow \phi = 100 \times \dfrac{1}{2} \times {10^{ - 4}} \\
\Rightarrow \phi = 5 \times {10^{ - 3}}\,{\text{Wb}} \\
\]

Hence, the magnetic field linked with the coil is \[5 \times {10^{ - 3}}\,{\text{Wb}}\] .
The correct option is A.

Note:To solve this problem, you should have some knowledge on the magnetic effects of current. Many students forget to convert the unit of the cross-sectional area to S.I units, which may produce undesirable results. Magnetic flux can be increased by making the field stronger, which we can do by placing a magnet near the coil.