A square loop of side 0.78 m is placed in a region where there is a magnetic fie
ID: 1731450 • Letter: A
Question
A square loop of side 0.78 m is placed in a region where there is a magnetic field of strength 18 T directed perpendicularly out of the plane of the loop If the magnetic field is decreased to 9 T in 0.04 seconds, If the resistance of the loop is 8 Ohm, determine the average current induced in the loop. (1 point) A. 27.368 A B. 29.488 A C. 10.516 A D. 4.14 A A square loop of side 0.89 m is placed in a region where there is a magnetic field of strength 14 T directed perpendicularly into the plane of the loop If the shape of the square is changed to a circle in 0.2 seconds, calculate the average emf induced in the loop. (1 point) A. 6.761 V B. 28.53 V C. 15.15 V D. O24.473 V E. 12.812V A square loop of side 0.78 m and resistance 6 Ohm is placed in a region where there is a magnetic field of strength 17 T plane of the loop. If the shape of the square is changed to a circle in 0.6 seconds, Determine the average current induced in the loop. (1 point) directed perpendicularly out of the A 0.148 A B. O1.165 A C. 1.05 A D. 0.785 A E. 1.295 AExplanation / Answer
Given
square loop of side a = 0.78 m
B1 = 18 T,B2 = 9 T
dt = 0.04 s
R = 8 ohm
induced current i =?
we know from Lenz's law induced emf e = -(dphi)/dt
phi is magnetic flux phi = B*A cos theta
theta is the angle between B and normal to the surface area here theta = 0
d phi = 9-18 = -9 T
Area of the square loop is A = a^2 = 0.78^2 m^2
e = -(dphi)/dt
e = -(A*dB)/dt
e = -A(dB/dt)
e = -0.78^2 (-9/0.04) V
e = 136.89 V
from Ohm's law V = I*R
I = V/R
I = 136.89/8 A
I = 17.11125 A
I = 17.111 A
the answer is option E 17.111 A
-----------------
given
square loop side a = 0.89 m
B = 14 T
dt =0.2 s
here the induced emf e = -dphi/dt
dphi =B*dA/dt
dA is change in area of the circle
first r1 radius of the circle which is equivalent to the 0.89 m side square area is
0.89^2 = pi*r1^2
r1 = 0.50213 m later
became circle which is of circumferance having within the square is
4*a = 2pi*r2
4*0.89 = 2pi*r2
r2 = 0.56659 m
now the change in area is dA = pi(r2^2-r1^2) = pi(0.56659^2 -0.50213 ^2)
e = 14pi(0.56659^2 -0.50213^2)/0.2 V
e = 15.1496 V
e = 15.15 V
answer is option C 15.15 V
--------------
given
square loop side a = 0.78 m
B = 17 T
dt =0.6 s
R = 6 ohm
here the induced emf e = -dphi/dt
dphi =B*dA/dt
dA is change in area of the circle
first r1 radius of the circle which is equivalent to the 0.89 m side square area is
0.78^2 = pi*r1^2
r1 = 0.4400 m later
became circle which is of circumferance having within the square is
4*a = 2pi*r2
4*0.78 = 2pi*r2
r2 = 0.4966 m
now the change in area is dA = pi(r2^2-r1^2) = pi(0.4966^2 -0.4400^2)
e = 17pi(0.4966^2 -0.4400^2)/0.6 V
e = 4.71865 V
from Ohm's law
I = e/R = 4.71865 /6 = 0.785 A
answer is option D , I = 0.785 A
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.