e 8 33% Part (b) Using the equation for the magnetic field determined in part (a), calculate the magnetic flux, in webers, through a single loop of the solenoid with a current of I=2.1 A. The coil is d= 6 cm long and has a cross-sectional area of A=4 cm² and consists of N=190 turns. 4 - 635.1 10-6 $= 6.351E-4 X Attempts Remain D 8 33% Part (c) The self inductance relates the magnetic flux linkage to the current through the coil. Calculate the self inductance L in units of µH. The coil 1s d = 6 cm long and has a cross-sectional area of A = 4 cm and consists of N= 190 turns. L =0.00030243| Grade Summary Deductions Potential 39 97% sin() cos) tan() HOME Submissions cotan) asin() E 1^ acos) sinh) 4 6 Attempts remaning: (39% per attempt) detailed viewv atan() acotan() 12 3 cosh() tanh() cotanh() END 306 O Degrees Radians VO BACKSPACE DEL CLEAR

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(10%) Problem 8: A solenoid consists of N= 190 turns of wire in a coil of length d = 6 cm and cross-sectional area A = 4 cm². Assume that
the magnetic field is uniform inside the solenoid and ignore end effects.
EV 33% Part (a) Write an equation for the magnetic field B produced by the solenoid. Express the answer in terms of the current through the coil I, the
number of turns N, length of the coil d, and the permeability of free space Ho-
B = H, NId Correct!
e 8 33% Part (b) Using the equation for the magnetic field determined in part (a), calculate the magnetic flux, in webers, through a single loop of the
solenoid with a current of I=2.1 A. The coil is d= 6 cm long and has a cross-sectional area of A=4 cm² and consists of N=190 turns.
$ = 635.1 * 10-6
$ = 6.351E-4
X Attempts Remain
D 8 33% Part (c) The self inductance relates the magnetic flux linkage to the current through the coil. Calculate the self inductance L in units of µH. The coil
1s d = 6 cm long and has a cross-sectional area of A = 4 cm² and consists of N= 190 turns.
L =0.00030243|
Grade Summary
Deductions
Potential
390
97%
sin()
cos()
tan()
7.
HOME
Submissions
cotan()
asin()
acos)
E 1^
4.
5
Attempts remaning: 1
(19% per attempt)
detailed view
atan()
acotan()
sinh()
123
cosh()
cotanh()
O Degrees O Radians
tanh()
END
1
300
Vol BACKSPACE
DEL
CLEAR
Transcribed Image Text:(10%) Problem 8: A solenoid consists of N= 190 turns of wire in a coil of length d = 6 cm and cross-sectional area A = 4 cm². Assume that the magnetic field is uniform inside the solenoid and ignore end effects. EV 33% Part (a) Write an equation for the magnetic field B produced by the solenoid. Express the answer in terms of the current through the coil I, the number of turns N, length of the coil d, and the permeability of free space Ho- B = H, NId Correct! e 8 33% Part (b) Using the equation for the magnetic field determined in part (a), calculate the magnetic flux, in webers, through a single loop of the solenoid with a current of I=2.1 A. The coil is d= 6 cm long and has a cross-sectional area of A=4 cm² and consists of N=190 turns. $ = 635.1 * 10-6 $ = 6.351E-4 X Attempts Remain D 8 33% Part (c) The self inductance relates the magnetic flux linkage to the current through the coil. Calculate the self inductance L in units of µH. The coil 1s d = 6 cm long and has a cross-sectional area of A = 4 cm² and consists of N= 190 turns. L =0.00030243| Grade Summary Deductions Potential 390 97% sin() cos() tan() 7. HOME Submissions cotan() asin() acos) E 1^ 4. 5 Attempts remaning: 1 (19% per attempt) detailed view atan() acotan() sinh() 123 cosh() cotanh() O Degrees O Radians tanh() END 1 300 Vol BACKSPACE DEL CLEAR
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