© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Key Parameters
• Number of turns in the coil, $N = 1000$.
• Average radius of the coil, $r = 10\, \text{m}$.
• Angular speed of rotation, $\omega = 2\, \text{rad s}^{-1}$.
• Vertical component of Earth’s magnetic field, $B = 2 \times 10^{-5}\, \text{T}$.
• Electrical resistance of the coil, $R = 12.56\, \Omega$.
Step 2: Write the Magnetic Flux through the Coil
The magnetic flux through one turn of the coil, assuming the axis of rotation is perpendicular to the direction of the given magnetic field (vertical component), can be written as:
$\phi_B = B \times A \times \cos(\omega t)$
Since there are $N$ turns, the total flux is:
$\phi_B = N \, B \, A \, \cos(\omega t)$
Here, $A$ is the area of the coil, given by:
$A = \pi r^2 = \pi \times (10)^2 = 100\pi \, \text{m}^2.$
Step 3: Find the Induced EMF
According to Faraday’s law of electromagnetic induction, the induced emf $ \varepsilon $ is:
$\varepsilon (t) = - \frac{d \phi_B}{dt}.$
For $\phi_B = N B A \cos(\omega t)$,
$\varepsilon (t) = - \frac{d}{dt} \bigl( N B A \cos(\omega t) \bigr) = N B A \omega \sin(\omega t).$
Thus, the maximum value of the induced emf occurs when $\sin(\omega t) = 1$, so
$\varepsilon_{\max} = N B A \omega.$
Step 4: Calculate the Maximum Induced Current
Ohm’s law relates current to voltage (emf) and resistance. Hence, the maximum induced current is:
$i_{\max} = \frac{\varepsilon_{\max}}{R} = \frac{N B A \omega}{R}.$
Substitute the values:
• $N = 1000$
• $B = 2 \times 10^{-5}\,\text{T}$
• $A = \pi \times (10)^2 = 100\pi \,\text{m}^2$
• $\omega = 2\,\text{rad s}^{-1}$
• $R = 12.56\,\Omega$
Therefore,
$i_{\max} = \frac{1000 \times (2 \times 10^{-5}) \times 100\pi \times 2}{12.56}.$
Step 5: Perform the Numerics
Calculate the numerator first:
$1000 \times 2 \times 10^{-5} \times 100\pi \times 2 = 1000 \times 2 \times 100 \times 2 \times \pi \times 10^{-5}.$
• $1000 \times 2 = 2000.$
• $2000 \times 100 = 200\,000.$
• $200\,000 \times 2 = 400\,000.$
• Factor in $10^{-5}$: $400\,000 \times 10^{-5} = 400\,000 \times 0.00001 = 4.$
• Finally, multiply by $\pi$: $4 \pi.$
So the numerator is approximately $4\pi.$
Now divide by $R = 12.56.$
$i_{\max} = \frac{4\pi}{12.56}.$
Numerically, $4\pi \approx 12.56,$ so
$i_{\max} \approx \frac{12.56}{12.56} = 1\, \text{A}.$
Step 6: Final Answer
Hence, the maximum induced current in the coil is 1 A.
