Question

A coil of area $5\,c{m^2}$ with 20 turns is kept under the magnetic field of ${10^3}\,gauss$. Normal to the plane of the coil makes an angle ${30^ \circ }$ with the magnetic field. What is the flux through the coil?
A. $6.67 \times {10^{ - 4\,}}\,Wb$
B. $3.2 \times {10^{ - 5}}\,Wb$
C. $5.9 \times {10^{ - 4\,}}\,Wb$
D. $8.65 \times {10^{ - 4\,}}\,Wb$

Answer 1

Hint: We know that the magnetic flux is the number of magnetic field lines passing through a given area. It depends directly on the strength of the magnetic field, area, and angle made by the magnetic field and the normal to the area. It also depends on the number of turns present in the coil.

Formula used:
$\phi = NBA\cos \theta $
Where $N$ is the number of turns of the coil, $B$ is the magnetic field, $A$ is the area and $\theta $ is the angle between the magnetic field and normal to the plane of the coil.

Complete step by step answer:
Flux through a coil is the number of magnetic field lines passing through a given area. It is denoted by the symbol $\phi $.
In the question area of the coil is given as
$A = 5\,c{m^2}$
$ \Rightarrow A = 5 \times {10^{ - 4\,}}{m^2}\,$
Number of turns of the coil is given as,
$N = 20$
Magnetic field is given as,
$B = {10^3}\,gauss$
$ \Rightarrow B = {10^3}\, \times {10^{ - 4}}\,T$
$\because 1\,T = {10^{4\,}}gauss$
$ \Rightarrow B = {10^{ - 1}}\,T$
Tesla is the SI unit of magnetic field where, $1\,T = 1\,Wb/{m^2}$
Angle between magnetic field and normal to the plane of the coil is given as ,
$\theta = {30^ \circ }$
We know that flux through a coil gives the total number of magnetic field lines passing through a given area. It depends upon the strength of the magnetic field, area and the angle made by the magnetic field and the normal to the area. It also depends on the number of turns present in the coil.
In equation form we can write it as,
$\phi = NBA\cos \theta $
Where $N$ is the number of turns of the coil.$B$ is the magnetic field $A$ is the area and $\theta $ is the angle between magnetic field and normal to the plane of the coil.
Now let us Substitute the given values in the equation.
Then we get,
$\phi = 20 \times {10^{ - 1}}\,T \times 5 \times {10^{ - 4\,}}{m^2}\, \times \cos {30^ \circ }$
$ \Rightarrow \phi = 8.65 \times {10^{ - 4\,}}\,Wb$

Therefore, the correct answer is option D.

Note:
The flux will be maximum when the angle between magnetic field and normal to the area is ${0^ \circ }$ since $\cos {0^ \circ } = 1$ and flux will be minimum when the angle is ${90^ \circ }$ since $\cos \,{90^ \circ } = 0$.
While doing this problem take care of the unit in which the magnetic field is given. The SI unit of the magnetic field is tesla, $T$. $1\,{\text{tesla}}$ is $1\,Wb/{m^2}$ Since it is given in gauss convert it into Tesla using the conversion formula $1\,T = {10^{4\,}}gauss$.