A solid cylindrical conducting shell carries a uniformly An infinite conducting wire carrying a current into the page is located at the center as shown. Which is a correct application of Ampere's law to solve for the magnetic field B around an Amperian loop of radius 8.0 cm centered on the center wire? Use the convention that current out of the screen is positive and current into the screen is negative, so that positive B is counterclockwise and negative B is clockwise. Take the units of B to be Tesla and μo to be in SI units as in the list of constants above.

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6.6A
O
2.64
Ⓒ4,2cm 2.64
ст
17cm
A solid cylindrical conducting shell carries a uniformly distributed current out of the page as shown.
An infinite conducting wire carrying a current into the page is located at the center as shown. Which
is a correct application of Ampere's law to solve for the magnetic field B around an Amperian loop of
radius 8.0 cm centered on the center wire? Use the convention that current out of the screen is
positive and current into the screen is negative, so that positive B is counterclockwise and negative
B is clockwise. Take the units of B to be Tesla and to be in SI units as in the list of constants above.
Ο Β 2 π (0.07)= + 4.0 μο
Ο Β 2 π (0.07)= -6.6 μο
Ο Β 2 π (0.08)= + 4.0 μο
Ο Β 2 π (0.08)= + 6.6 μο
Ο Β 2 π (0.08)= - 4.0 μο
Transcribed Image Text:6.6A O 2.64 Ⓒ4,2cm 2.64 ст 17cm A solid cylindrical conducting shell carries a uniformly distributed current out of the page as shown. An infinite conducting wire carrying a current into the page is located at the center as shown. Which is a correct application of Ampere's law to solve for the magnetic field B around an Amperian loop of radius 8.0 cm centered on the center wire? Use the convention that current out of the screen is positive and current into the screen is negative, so that positive B is counterclockwise and negative B is clockwise. Take the units of B to be Tesla and to be in SI units as in the list of constants above. Ο Β 2 π (0.07)= + 4.0 μο Ο Β 2 π (0.07)= -6.6 μο Ο Β 2 π (0.08)= + 4.0 μο Ο Β 2 π (0.08)= + 6.6 μο Ο Β 2 π (0.08)= - 4.0 μο
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