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
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