A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

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A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

orm magnetic field given by B⃗ =(1.05×10−2T)(12i^+3j^−4k^). The ring is positioned initially such that its magnetic moment orientation is given by μ⃗ i=μ(−0.8i^+0.6j^), where μ is the (positive) magnitude of the magnetic moment. (a) Find the initial magnetic torque on the ring. (b) The ring (which is free to rotate around one diameter) is released and turns through an angle of 90.0∘, at which point its magnetic moment orientation is given by μ⃗ f=−μk^. Determine the decrease in potential energy. (c) If the moment of inertia of the ring about a diameter 6.50×10−7kg⋅cm2, determine the angular speed of the ring as it passes through the second position.

Physics

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-03-09T12:47:16+03:00

a) (0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^) N.m b) ΔU = -0.000747871 J c) w = 47.97 rad/s The relevant data includes the area of the circular ring as 4.45 cm², the current of 13.5 Amps, and the magnetic field strength as (1.05×10−2T)(12i^ + 3j^ - 4k^). The initial magnetic moment orientation is expressed as μi = μ(−0.8i^ + 0.6j^). The moment of inertia for the ring is given as 6.50×10−7 kg⋅m². To find the initial magnetic moment (μ), we apply μ = N*I*A, using N=1 for a single coil. Substituting the values yields μ = 0.0060075 A-m². The torque on the ring is determined using the cross product of the magnetic moment and the magnetic field. Calculating this gives the initial torque as (0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^). The magnetic potential energy is then calculated using the dot product of the magnetic moment with the magnetic field. This leads to initial energy stored in the ring as Ui = 0.000495556 J after computation. Following a 90-degree rotation, the final potential energy is found to be Uf = -0.000252315 J. The decrease in potential energy is determined to be ΔU = -0.000747871 J. Thus, the potential energy stored in the ring decreased by ΔU = -0.000747871 J.