Question 15: Electromagnetic Induction

Question

A wire ' 1 ' $$\mathrm{cm}$$ long bent into a circular loop is placed perpendicular to the magnetic field of flux density '$$B^{\prime} W b \mathrm{~m}^{-2}$$. Within $$0.1 \mathrm{sec}$$, the loop is changed into a square of side '$$a$$' $$\mathrm{cm}$$ and flux density is doubled. The value of e.m.f. induced is

Accepted Answer

We start by noting that the induced emf (by FaradayÃ¢ÂÂs law) is given by
$$ \varepsilon = -\frac{\Delta\Phi}{\Delta t}, $$
where the flux change
$$ \Delta\Phi = \Phi_{	ext{final}} - \Phi_{	ext{initial}}. $$
In our problem the magnetic field is perpendicular to the loop (so we donÃ¢ÂÂt have to worry about any angle factors) and its value doubles when the loop is changed in 0.1 sec. LetÃ¢ÂÂs set up the steps:
1.Ã¢ÂÂDefine l as the length of the wire (in cm) and a as the side of the square (in cm) into which the wire is reformed. (In many problems the wire length l is used; here the given value Ã¢ÂÂ1 cmÃ¢ÂÂ is simply a particular case, so we write our answer in terms of l and a.)
2.Ã¢ÂÂWhen the wire is bent into a circle, its circumference is l. That means the radius r of the circle is
$$ r = \frac{l}{2\pi}. $$
Therefore, the area of the circular loop is
$$ A_{	ext{circle}} = \pi r^2 = \pi\left(\frac{l}{2\pi}ight)^2 = \frac{l^2}{4\pi}. $$
3.Ã¢ÂÂWhen the loop is changed into a square of side a, its area is simply
$$ A_{	ext{square}} = a^2. $$
(Note: In a typical wireÃ¢ÂÂreshaping problem the new shapeÃ¢ÂÂs perimeter would equal l so that a would be l/4. Here, however, the side a is given explicitly, so we use it as an independent parameter.)
4.Ã¢ÂÂThe initial magnetic flux is that through the circular loop in a field B:
$$ \Phi_{	ext{initial}} = B\,A_{	ext{circle}} = B\,\frac{l^2}{4\pi}. $$
After 0.1 sec the loop is made into a square and, at the same time, the magnetic flux density is doubled to 2B. Therefore the final flux is
$$ \Phi_{	ext{final}} = (2B)\,A_{	ext{square}} = 2B\,a^2. $$
5.Ã¢ÂÂThus the change in flux is
$$ \Delta\Phi = \Phi_{	ext{final}} - \Phi_{	ext{initial}} = 2B\,a^2 - B\,\frac{l^2}{4\pi}. $$
6.Ã¢ÂÂPlug this into FaradayÃ¢ÂÂs law with the time interval $ \Delta t = 0.1\,	ext{sec} $:
$$ \varepsilon = -\frac{\Delta\Phi}{\Delta t}  = -\frac{2B\,a^2 - \dfrac{B\,l^2}{4\pi}}{0.1}. $$
Multiplying numerator and denominator by 10 (since $1/0.1 = 10$) gives
$$ \varepsilon = -10B\left(2a^2 - \frac{l^2}{4\pi}ight). $$
7.Ã¢ÂÂTo combine the terms inside the parentheses, rewrite $2a^2$ with a common denominator:
$$ 2a^2 = \frac{8\pi\,a^2}{4\pi}. $$
Thus we have
$$ \varepsilon = -10B\,\left(\frac{8\pi\,a^2 - l^2}{4\pi}ight) = -\frac{10B}{4\pi}\,(8\pi\,a^2 - l^2). $$
Simplify the factor $\frac{10}{4} = \frac{5}{2}$, so
$$ \varepsilon = -\frac{5B}{2\pi}\,(8\pi\,a^2 - l^2). $$
This is exactly the expression given in option C:
$$ 	extbf{Option C:}\quad \varepsilon = \frac{-5B\,(8\pi a^2 - l^2)}{2\pi}. $$
Thus, the induced emf is
$$ \varepsilon = -\frac{5B\,(8\pi\,a^2 - l^2)}{2\pi}, $$
and the correct answer is Option C.

Answer

$$
\frac{-5 B\left(8 л \mathrm{a}^2-\mathrm{l}^2\right)}{2}
$$

Answer

$$
\frac{5 B\left(8 л a^2-l^2\right)}{2 л}
$$

Answer

$$
\frac{-5 B\left(8 л a^2-l^2\right)}{2 л}
$$

Answer

$$
\frac{-5 B\left(8 a^2-l^2\right)}{2 \pi}
$$

Electromagnetic Induction

A wire ' 1 ' $\mathrm{cm} long bent into a circular loop is placed perpendicular to the magnetic field of flux density 'B^{\prime} W b \mathrm{~m}^{-2}. Within 0.1 \mathrm{sec}, the loop is changed into a square of side 'a' \mathrm{cm}$ and flux density is doubled. The value of e.m.f. induced is

Choose the correct answer: