Respuesta :
Answer:
Thus induced emf is 0.0531 V
Solution:
As per the question:
Diameter of the loop, [tex]d = 9.6\ cm = 0.096\ m[/tex]
Thus the radius of the loop, R = 0.048 m
Time in which the loop is removed, t = 0.15 s
Magnetic field, B = 1.10 T
Now,
The average induced emf, e is given by Lenz Law:
[tex]e = - \frac{\Delta \phi_{B}}{\Delta t}[/tex]
[tex]e = - \frac{\Delta \phi_{B}}{\Delta t}[/tex]
where
[tex]\phi_{B}[/tex] = magnetic flux = [tex]A\Delta B[/tex]
where
A = cross sectional area
Also, we know that:
[tex]e = - \frac{A\Delta B}{\Delta t}[/tex]
[tex]e = - \frac{\pi r^{2}\times (0 - 1.10)}{0.15}[/tex]
[tex]e = - \frac{\pi \times 0.048^{2}\times (0 - 1.10)}{0.15}[/tex]
e = 0.0531 V
The sketch is shown in the figure, where I indicates the direction of the induced current.

Answer: [tex]emf=0.531 V[/tex]
Explanation:
Induced EMF is equals to the change in the Magnetic flux per unit change in time.
It is calculated by the Faraday's law which is given mathematically as:
[tex]emf=N \times \frac{\Delta (B.A) }{\Delta t}[/tex] ..................................(1)
where:
- N= number of turns in looped coil of wire
- B= magnetic field
- A= area of the loop formed by the wire
- t= change in time
- [tex] \Delta [/tex]= change in the quantity from initial to final state.
Given:
N=1 (since it is a loop not a coil of loops)
Diameter of loop, d = 9.6 cm
[tex]B = 1.10 T[/tex]
[tex]{\Delta t}[/tex]= 0.15 s
Firstly, we find the area of loop:
[tex]A= \pi \frac{d^{2}}{4}[/tex]
[tex]A= \pi \frac{(9.6 \times 10^{-2} )^{2}}{4}[/tex]
[tex]A= 7.2382 \times 10^{-3} m^{2}[/tex]
Putting respective values in eq. (1)
[tex]emf=1\times \frac{\Delta (1.10\times 7.2382 \times 10^{-3}) }{0.15}[/tex]
[tex]emf=0.531 V[/tex]
THe direction of the induced emf is in accordance with the Lenz Law which states that the direction of an induced current is always such as to oppose the cause by which it is produced.
