Answer:
The average induced emf in the loop is 0.20 V
Explanation:
Given:
Radius of loop [tex]r = \frac{d}{2} = 9.25 \times 10^{-2}[/tex] m
Magnetic field [tex]B = 1.5[/tex] T
Change in time [tex]\Delta t = 0.20[/tex] sec
According to the faraday's law,
Induced emf is given by
[tex]\epsilon = -\frac{\Delta \phi}{\Delta t}[/tex]
Where [tex]\phi =[/tex] magnetic flux
[tex]\phi = BA\cos0[/tex] ( here [tex]\theta = 0[/tex] )
Where [tex]A = \pi r^{2}[/tex]
We neglect minus sign because it's shows lenz law
[tex]\epsilon = \frac{B \pi r^{2} }{\Delta t}[/tex]
[tex]\epsilon = \frac{1.5 \times 3.14 \times (9.25 \times 10^{-2} )^{2} }{0.20}[/tex]
[tex]\epsilon = 0.20[/tex] V
Therefore, the average induced emf in the loop is 0.20 V
The average induced emf in the loop will be "0.20 V". To understand the calculation, check below.
According to the question,
Diameter, d = 18.5 cm
Radius, r = [tex]\frac{d}{2}[/tex]
= [tex]\frac{18.5}{2}[/tex]
= 9.25 × 10⁻² m
Magnetic field, B = 1.5 T
Change in time, Δt = 0.20 sec
According to Faraday's law,
→ [tex]\epsilon[/tex] = [tex]-\frac{\Delta \phi}{\Delta t}[/tex]
or,
Magnetic flux, [tex]\phi[/tex] = BA Cos0
hence,
The induced emf be:
→ [tex]\epsilon[/tex] = [tex]\frac{B \pi r^2}{\Delta t}[/tex]
By substituting the values,
= [tex]\frac{1.5\times 3.14\times (9.25\times 10^{-2})^2}{0.20}[/tex]
= 0.20 V
Thus the above answer is correct.
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