A loop of diameter d = 12 cm, carrying a current I = 0.4 A is placed inside a ma

Question

A loop of diameter d = 12 cm, carrying a current I = 0.4 A is placed inside a magnetic field B = (0.2 T)i + (0.4)j. The normal to the loop is parallel to the unit vector a = 0.6i-0.8j. What is the otential energy of the loop? A) -4.5 times 10^-4 J B) +9.0 times 10^-4 J C) -9.0 times 10^-4 J D) -2.3 times 10^-4 J E) +4.5 times 10^-4 J 6. What is the direction of the induced current in R when the currebt I in the figure decrease rapidly to zero? a) Undeterminable from the information given b) From b to a through R c) From a to b through R 7. A hollow cylinder of inner radius a = 5 cm and outer radius b = 7 cm. A uniform current density j = 1 A/cm^2 flows through the cylinder. Calculate the magnitude of magnetic field at a distance d = 10 cm from the axis of the cylinder. A) 0 B) 1.5 times 10^-4 T C) 2.5 times 10^-4 T D) 4.5 times 10^-4 T E) 0.5 times 10^-4 T 8. A cooper bar moved to the right while its axis is maintained perpendicular to a magnetic field, as in the figure. If the top of the bar becomes positive relative to the bottom, what it the direction of the magnetic field? b) Directed into the paper b) Directed out of the paper c) Undeterminable from the information given

Explanation / Answer

5) diameter of loop (d) = 0.12 m

radius of loop (r) = 0.06 m

I = 0.40 A

B = 0.20 Ti + 0.40 T j

n = 0.60i - 0.80 j

Then potential energy associated with the magnetic moment = U = - u.B

u = IAn = 0.40 A * 3.14 * (0.06)^2 * (0.60i - 0.80 j)

=> U = - u.B

=> U = - 0.40 A * 3.14 * (0.06)^2 * (0.60i - 0.80 j). [0.20 Ti + 0.40 T j]

=> U = - 0.0045216 [0.12 - 0.32]

=> U = 0.00090432 J = +9*10^-4 J (option B)

6) From b to a through R. (option B) using Lenz law

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