Math  /  Algebra

QuestionCool-down The LL-shaped conductor in the figure below moves at 10 m/s10 \mathrm{~m} / \mathrm{s} across and touches a stationary LL-shaped conductor in a 0.1 T magnetic field. The two vertices overlap, so that the enclosed area is zero, at t=0t=0. The conductor has a resistance of 0.02Ω0.02 \Omega per meter. V=10 mV=10 \mathrm{~m} i. What is the direction of the induced current? ii. Find expressions for the induced emf and the induced current as functions of time. iii. Evaluate the induced emf and current at t=0.1 st=0.1 \mathrm{~s}.

Studdy Solution
Evaluate the induced emf and current at t=0.1st = 0.1 \, \text{s}.
Substitute t=0.1t = 0.1 into the expression for the area:
A=100×0.1=10m2A = 100 \times 0.1 = 10 \, \text{m}^2
The induced emf remains constant at 10V-10 \, \text{V} since it depends on the rate of change of area, not the area itself.
The induced current is:
I=25AI = -25 \, \text{A}
The induced emf at t=0.1st = 0.1 \, \text{s} is 10V-10 \, \text{V} and the induced current is 25A-25 \, \text{A}.

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