Key Concept: Complex Induced Emf Scenario, Advanced Lorentz Force Scenario
c) 8.94 mV
[Solution Description]
To solve this problem, we need to find the change in magnetic flux through the loop as it rotates and then use Faraday’s law to calculate the induced emf.
The area of the loop is:
$$A = 5 \times 10 = 50 \, \text{cm}^2 = 50 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-3} \, \text{m}^2$$
The initial magnetic flux through the coil can be determined considering that initially the plane of the loop is perpendicular to the magnetic field:
$$\Phi_{B_i} = B_i \cdot A \cdot \cos(\theta_i) = 0$$
Since the loop starts rotating from rest, $$\theta_i = 90^\circ$$, making the cosine term zero.
After rotating through $$45^\circ$$, the orientation of the loop changes, the new angle between magnetic field and area vector becomes $$\theta_f = 45^\circ$$. Therefore, the flux is:
$$\Phi_{B_f} = (kx \cdot l)(w \cdot h)\cos(45^\circ)$$
The rate of change of flux can be modeled as:
$$\frac{d\Phi_B}{dt} = \frac{\Phi_{B_f} – \Phi_{B_i}}{\Delta t}$$
Since the transition takes time related to its angular speed:
$$\Delta t = \frac{45^\circ}{\omega} = \frac{\pi/4}{100} = \frac{\pi}{400} \, \text{s}$$
Using Faraday’s Law for absolute value:
$$|\epsilon| = \left| – \frac{d\Phi_B}{dt} \right|$$
We substitute known values into the equation and get the result:
$$|\epsilon| = \left| – \frac{1 \times 10^{-3} \times 5 \times 10^{-3} \times \sin(45^\circ)}{\pi / 400} \right|$$
Simplifying further:
$$|\epsilon| = \frac{5 \times 10^{-6} \times \sqrt{2}/2}{\pi / 400}$$
Calculating the above expression gives:
$$|\epsilon| \approx 8.94 \, \text{mV}$$
Therefore, the magnitude of the induced emf is approximately $$8.94 \, \text{mV}.$$
Your Answer is correct.
c) 8.94 mV
[Solution Description]
To solve this problem, we need to find the change in magnetic flux through the loop as it rotates and then use Faraday’s law to calculate the induced emf.
The area of the loop is:
$$A = 5 \times 10 = 50 \, \text{cm}^2 = 50 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-3} \, \text{m}^2$$
The initial magnetic flux through the coil can be determined considering that initially the plane of the loop is perpendicular to the magnetic field:
$$\Phi_{B_i} = B_i \cdot A \cdot \cos(\theta_i) = 0$$
Since the loop starts rotating from rest, $$\theta_i = 90^\circ$$, making the cosine term zero.
After rotating through $$45^\circ$$, the orientation of the loop changes, the new angle between magnetic field and area vector becomes $$\theta_f = 45^\circ$$. Therefore, the flux is:
$$\Phi_{B_f} = (kx \cdot l)(w \cdot h)\cos(45^\circ)$$
The rate of change of flux can be modeled as:
$$\frac{d\Phi_B}{dt} = \frac{\Phi_{B_f} – \Phi_{B_i}}{\Delta t}$$
Since the transition takes time related to its angular speed:
$$\Delta t = \frac{45^\circ}{\omega} = \frac{\pi/4}{100} = \frac{\pi}{400} \, \text{s}$$
Using Faraday’s Law for absolute value:
$$|\epsilon| = \left| – \frac{d\Phi_B}{dt} \right|$$
We substitute known values into the equation and get the result:
$$|\epsilon| = \left| – \frac{1 \times 10^{-3} \times 5 \times 10^{-3} \times \sin(45^\circ)}{\pi / 400} \right|$$
Simplifying further:
$$|\epsilon| = \frac{5 \times 10^{-6} \times \sqrt{2}/2}{\pi / 400}$$
Calculating the above expression gives:
$$|\epsilon| \approx 8.94 \, \text{mV}$$
Therefore, the magnitude of the induced emf is approximately $$8.94 \, \text{mV}.$$