A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current `I=I_(0)cos(omegat)`. The emf induced in the smaller inner loop is nearly ?

Options

A. `pimu_(0)I_(0)(a^(2))/(b)omegasin(omegat)`
B. `(pimu_(0)I_(0))/(2)*(a^(2))/(b)omegacos(omegat)`
C. `(pimu_(0)I_(0)b^(2))/(a)omegacos(omegat)`
D. `(pimu_(0)I_(0))/(2)*(a^(2))/(b)omegasin(omegat)`

Answer

To find the induced EMF in the smaller circular loop due to the alternating current in the larger loop, we follow these steps:

### Step 1: Determine the Magnetic Field at the Center of the Larger Loop
The magnetic field \( B \) at the center of a circular loop of radius \( b \) carrying a current \( I \) is given by the formula:

\[ B = \frac{\mu_0 I}{2b} \]

where \( \mu_0 \) is the permeability of free space.

### Step 2: Substitute the Expression for Alternating Current
The current in the larger loop is given as \( I = I_0 \cos(\omega t) \). Substituting this into the magnetic field expression, we have:

\[ B = \frac{\mu_0 I_0 \cos(\omega t)}{2b} \]

### Step 3: Calculate the Magnetic Flux through the Smaller Loop
The magnetic flux \( \Phi \) through the smaller loop of radius \( a \) is given by:

\[ \Phi = B \cdot A \]

where \( A \) is the area of the smaller loop, \( A = \pi a^2 \). Thus, substituting for \( B \):

\[ \Phi = \left(\frac{\mu_0 I_0 \cos(\omega t)}{2b}\right) \cdot \pi a^2 \]

### Step 4: Differentiate the Magnetic Flux to Find Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is given by:

\[ \mathcal{E} = -\frac{d\Phi}{dt} \]

Differentiating the expression for \( \Phi \):

\[ \mathcal{E} = -\frac{d}{dt} \left(\frac{\mu_0 I_0 \cos(\omega t) \pi a^2}{2b}\right) \]

### Step 5: Apply the Chain Rule
Using the chain rule to differentiate \( \cos(\omega t) \):

\[ \frac{d}{dt} \cos(\omega t) = -\omega \sin(\omega t) \]

Thus, we have:

\[ \mathcal{E} = -\left(\frac{\mu_0 \pi a^2 I_0}{2b}\right)(-\omega \sin(\omega t)) \]

This simplifies to:

\[ \mathcal{E} = \frac{\mu_0 \pi a^2 I_0 \omega}{2b} \sin(\omega t) \]

### Final Expression for Induced EMF
The induced EMF in the smaller loop is:

\[ \mathcal{E} = \frac{\mu_0 \pi a^2 I_0 \omega}{2b} \sin(\omega t) \]