Question

A square coil of side $10{\text{ }}cm$ having $200$ turns is placed in a magnetic field of $1{\text{ W/}}{{\text{m}}^2}$ such that its plane is perpendicular to the field. The coil is rotated by $180^\circ$ in $0.5$ seconds. What is the induced emf?

Answer 1

These sums are based on basic formulas of Magnetic induction and questions asked are only formula based. We know from Faraday’s law of induction that emf= $N\dfrac{{d\phi }}{{dt}}$ , where $\phi$ = $B.A$, where $B$ is the magnetic field intensity and $A$ is the area of the coil. Thus, all the values are given, we need to obtain it by simple arithmetic.

Complete step by step answer:
Now we are provided with the following sets of data:
N= number of turns of coil, and $\dfrac{{d\phi }}{{dt}}$ = rate of change of flux.
First, we will find $\phi$. We know,
$\phi$ = $B.A$ $= 1 \times {\text{ }}{10^2} \times {10^{ - 4}}{\text{ }}$= $= {10^{ - 2}}$ $W$.
Hence we get $\phi$, now emf = $N\dfrac{{d\phi }}{{dt}}$
where $N$= 200, hence emf $= {\text{ }}\dfrac{{{\text{(}}200 \times {{10}^{ - 2}})}}{{0.5}}$

But here we see that the coil is being rotated through an angle of $180^\circ$ (Note this point as students often make mistakes here and consider only a single rotation). Hence the field is at first reduced to zero, angle changes: $0 - 90^\circ$ and then from zero to max, angle changes as: $90^\circ - 180^\circ$.
Thus, the resultant emf equals, applying the formula of emf:
$2\times \dfrac{{{\text{(}}200 \times {{10}^{ - 2}})}}{{0.5}}$.
Thus, emf = $8{\text{ V}}$. Thus, we have to read the question and get to the point and understand what it has to say with the angle to avoid mistakes.

Note: For these types of questions, remember that the angle and number of turns always play a major role. Also be careful with the units, for instance here the sides of a square are given in cm, whereas the value of field intensity is in square metres. Always make sure the units are matching and the values are correctly taken in the question.