Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
If the process is smooth and only takes the given time, the magnitude of the emf induced in the loop is 3.33 × 10⁻⁴Volts.
Given the data in the question
- Magnetic field; [tex]B = 20mT = 20 * 10^{-3}T[/tex]
- Area; [tex]A = 200cm^2 = 0.02m^2[/tex]
- Time; [tex]\delta t = 1.2s[/tex]
Magnitude of the emf induced in the loop; [tex]|e| =\ ?[/tex]
From Faraday's law, formula for Induced emf is given as:
[tex]|e| = -\frac{\delta\theta}{\delta t} \\\\|e| = -\frac{\delta (BAcos\theta)}{\delta t} \\\\|e| = -BA\frac{\delta (cos\theta)}{\delta t}[/tex]
Where [tex]\delta\theta[/tex] is the change in flux, [tex]\delta t[/tex] is time taken to change the flux, B is the Magnetic field and A is the area.
We know that; [tex]\theta = 0^o\ to\ 90^o[/tex], so
[tex]|e| = -BA\frac{(cos90^o -cos0^o)}{\delta t}[/tex]
Now, we substitute in our given valueslux
[tex]|e| = [-(20*10^{-3}T)*0.02m^2]\frac{(cos90^o -cos0^o)}{1.2s}\\\\|e| = (-0.0004Tm^2)\frac{(0-1)}{1.2s}\\\\|e| = \frac{0.0004Tm^2}{1.2s}\\\\|e| = 3.33*10^{-4}Volts[/tex]
Therefore, if the process is smooth and only takes the given time, the magnitude of the emf induced in the loop is 3.33 × 10⁻⁴Volts
Learn more: https://brainly.com/question/17110955
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.