10 2018 Ch 30,31 Begin Date: 3/28/2018 12:01:00 AM -- Due Date: 4/4/2018 11:59:0
ID: 2035464 • Letter: 1
Question
10 2018 Ch 30,31 Begin Date: 3/28/2018 12:01:00 AM -- Due Date: 4/4/2018 11:59:00 PM End Date: 4/4/2018 11:59:00 PM (17%) Problem 2: An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.00 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.1 cm and it takes 0.375 s to move it into the field Randomized Variables d-2.1 cm 0.375s ?? 33% Part (a) What average current is induced in the ring in A if its resistance is 0.0100 ?? 33% Part (b) What average power is dissipated in mw? Grade Sur Deductiorn Potential P (0.711 (117|819|HOME 1 | Submissio Attempts re (1% per at detailed vi tan() exs cotan0 sin cs atan acotan sinh0 cosh tnhotanh Sin 789HOME 4 5 6 0 END °Degrees Radians VOI BACKSPACE | DELI CLEAR 4 Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining:- Feedback: 0% deduction per feedback. 33% Part (c) What magnetic field is induced at the center of the ring in T?Explanation / Answer
given
problem 2: change in magnetic field = B2 - B1 = 2 T = dB
fingers pointing in the ddireciton of the field
diameter of wedding ring, d = 2.1 cm = 2.1*10^-2 m
time dt = 0.375 s
a. induced emf, from faradays law = dB*pi*d^2/4*dt = 2*pi(2.1*10^-2)^2/4*0.375 = 1.84725648 mV
as R = 0.01 ohm
hence
form ohms law
induced curret, i = V/R = 0.184725648031 A
b. average power dissipated = Vi = 0.184725648031 *1.84725648 = 0.3412356503 mW
c. magnetic field induced at the center of the ring = Bz
Bz = 4*pi*10^-7*i/d = 1.10539*10^-5 T
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