A coil of circular cross-section having 1000 turns and 4 cm$$^2$$ face area is placed with its axis parallel to a magnetic field which decreases by $$10^{-2}$$ Wb m$$^{-2}$$ in 0.01 s. The e.m.f. induced in the coil is:

To find the induced electromotive force (e.m.f.) in the coil, we use Faraday's law of electromagnetic induction. The law states that the induced e.m.f. is given by the formula:

$$ \text{e.m.f.} = -N \frac{d\phi}{dt} $$

where $$ N $$ is the number of turns in the coil, and $$ \frac{d\phi}{dt} $$ is the rate of change of magnetic flux through one turn. The negative sign indicates direction (Lenz's law), but for magnitude, we take the absolute value.

The magnetic flux $$ \phi $$ for one turn is $$ \phi = B \cdot A $$, where $$ B $$ is the magnetic field strength and $$ A $$ is the area perpendicular to the field. The coil's axis is parallel to the magnetic field, meaning the field is perpendicular to the coil's face, so this formula applies directly.

Given:

• Number of turns, $$ N = 1000 $$
• Face area, $$ A = 4 \text{cm}^2 $$

Convert area to square meters (since magnetic field is in Wb m$$^{-2}$$):

$$ 4 \text{cm}^2 = 4 \times 10^{-4} \text{m}^2 \quad \text{(because } 1 \text{cm}^2 = 10^{-4} \text{m}^2\text{)} $$

The magnetic field decreases by $$ 10^{-2} \text{Wb m}^{-2} $$ in $$ 0.01 \text{s} $$. So, the change in magnetic field $$ \Delta B = -10^{-2} \text{Wb m}^{-2} $$ (negative for decrease), and the time interval $$ \Delta t = 0.01 \text{s} $$.

The rate of change of magnetic field is:

$$ \frac{\Delta B}{\Delta t} = \frac{-10^{-2}}{0.01} = \frac{-0.01}{0.01} = -1 \text{Wb m}^{-2} \text{s}^{-1} $$

The magnitude of this rate is $$ \left| \frac{\Delta B}{\Delta t} \right| = 1 \text{Wb m}^{-2} \text{s}^{-1} $$.

For one turn, the rate of change of flux is:

$$ \frac{\Delta \phi}{\Delta t} = A \cdot \frac{\Delta B}{\Delta t} $$

Substituting the values:

$$ \frac{\Delta \phi}{\Delta t} = (4 \times 10^{-4}) \times (-1) = -4 \times 10^{-4} \text{Wb/s} $$

The magnitude is $$ \left| \frac{\Delta \phi}{\Delta t} \right| = 4 \times 10^{-4} \text{Wb/s} $$.

For the entire coil with $$ N $$ turns, the magnitude of the induced e.m.f. is:

$$ |\text{e.m.f.}| = N \times \left| \frac{\Delta \phi}{\Delta t} \right| = 1000 \times (4 \times 10^{-4}) $$

Calculate step by step:

$$ 1000 \times 4 \times 10^{-4} = 1000 \times 0.0004 = 0.4 \text{V} $$

Convert volts to millivolts (1 V = 1000 mV):

$$ 0.4 \text{V} = 0.4 \times 1000 \text{mV} = 400 \text{mV} $$

Alternatively, using the formula directly for magnitude:

$$ |\text{e.m.f.}| = N \cdot A \cdot \left| \frac{\Delta B}{\Delta t} \right| = 1000 \times (4 \times 10^{-4}) \times 1 = 0.4 \text{V} = 400 \text{mV} $$

Comparing with the options:

• A. 400 mV
• B. 200 mV
• C. 4 mV
• D. 0.4 mV

Hence, the correct answer is Option A.