A solenoid with 1200 turns per meter has a diameter of 5.00 cm. A current I = 2.
ID: 1403740 • Letter: A
Question
A solenoid with 1200 turns per meter has a diameter of 5.00 cm. A current I = 2.59 A flows in the clockwise direction in the solenoid. A rectangular loop of length L = 16.0 cm, width w = 12.5 cm, and 2 turns is centered on the axis of the solenoid.
(a) Find the magnitude of the magnetic flux through the loop.
(b) When the current is increased to 5.45 A, the magnitude of the induced emf in the rectangular loop is 116 mV. How long did it take for the current to get to this value?
ms
(c) What is the direction of the induced current in the rectangular loop as viewed from the location P?
counterclockwise, clockwise, or no current
Explanation / Answer
solenoid with 1200 turns per meter has a diameter of 5.00 cm. A current I = 2.59 A flows in the clockwise direction in the solenoid. A rectangular loop of length L = 16.0 cm, width w = 12.5 cm, and 2 turns is centered on the axis of the solenoid.
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(a) Find the magnitude of the magnetic flux through the loop.
(b) When the current is increased to 5.45 A, the magnitude of the induced emf in the rectangular loop is 116 mV. How long did it take for the current to get to this value?
ms
(c) What is the direction of the induced current in the rectangular loop as viewed from the location P?
counterclockwise, clockwise, or no current
Here ,
a)
number of turns , N = 1200
I = 2.59 A
magnetic field inside a loop , B = u0*N*I/L
B = 4pi*10^-7 * 1200*2.59/1
B = 3.905 *10^-3 T
B = 3.905 mT
magnetic flux through the loop = B*pi*r^2 * N
magnetic flux through the loop = 3.905 *10^-3 * 3.141 * 0.025^2 * 2
magnetic flux through the loop = 1.533 *10^-5 T.m^2
the magnetic flux through the loop is 1.533 *10^-5 T.m^2
b)
when the current is 5.45 A ,
flux in the loop = 4pi*10^-7 * 1200*5.45/1 * 3.141 * 0.025^2 * 2
flux in the loop = 3.227 *10^-5 T.m^2
let the time is t
0.116 = (3.227 - 1.533) *10^-5/t
t = 0.146 ms
the time taken is 0.146 ms
c)as the current in the loop is increasing ,
the current in the loop will be oposite to the current in the solenoid
the current in the loop is counterclockwise
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