# Induced EMF due to induced current.

1 vote
351 views

A very small circular loop of radius $a$ is initially coplanar and concentric with a much larger circular loop of radius $b$. A constant current $i$ is passed in the large loop which is kept fixed in space and the small loop is rotated with constant angular velocity $\omega$ about its diameter. The resistance of the small loop is $R$ and negligible $L$.

The induced emf in the large loop due to current induced in a smaller loop as a function of time is equal to $${\dfrac{1}{x}}\cdot \bigg(\dfrac{\pi a^2\mu_{o}\omega}{b}\bigg)^2\dfrac{i\cos2\omega t}{R}$$Find out $x$

For factor $\bigg(\dfrac{\pi a^2\mu_o\omega}{b}\bigg)^2$, I've got $\dfrac{(\pi a^2\mu_o\omega)^2}{4ab}$, not $b^2$.

I got, $i_a=\dfrac{\omega\mu_o i}{2bR}\cdot\pi a^2\sin\omega t$.

Now this will produce flux change through bigger loop,
$$\phi_b=\dfrac{\mu_o i_a}{2a}\cdot\pi a^2\cos\omega t=\dfrac{\mu_o}{2a}\cdot\bigg(\dfrac{\omega\mu_o i}{2bR}\cdot\pi a^2\sin\omega t\bigg)\cdot\pi a^2\cos\omega t$$ $$e_b={\dfrac{1}{4}}\cdot \dfrac{(\pi a^2\mu_o \omega)^2}{ab}\cdot\dfrac{i\cos2\omega t}{R}$$

asked Jun 29, 2018
edited Jun 30, 2018
The MathJax Editor did not like your brackets when you typed \mu_{o}. Brackets are not required for a single character. The Editor prefers \mu_o.

## 1 Answer

1 vote

Best answer

The mistake is in your expression for $\phi_b$. You have $ab$ in the denominator of $e_b$ instead of $b^2$.

The flux through small loop $a$ due to current $i_b$ in large loop $b$ can be written as $\phi_a=M(t) i_b$ where $M(t)$ is the time-dependent mutual inductance. Using the Reciprocity Theorem the flux through the large loop due to current $i_a$ in the small loop can be written $\phi_b=M(t) i_a$ See http://galileoandeinstein.physics.virginia.edu/2415/Physics_2415_Lec_22.pdf

You got $$\phi_a=\frac{\mu_0 i_b}{2b} \pi a^2 \cos\omega t=M(t) i_b$$ $$M(t)=\frac{\mu_0 \pi a^2 \cos\omega t}{2b}$$ So $a$ which you wrote in the denominator for $\phi_b$ should be $b$. You should have $$\phi_b=M(t) i_a=\frac{\mu_0 \pi a^2 \cos\omega t}{2b} i_a$$

answered Jun 30, 2018 by (28,466 points)
selected Jul 2, 2018 by n3