20 ohms in parallel with 16 ohm= 8.89
20x16/20+16. Product over sum
The solution to the questions are given as


- the direction of induced current will be Counterclock vise.
<h3>What is the direction of the
current induced in the loop, as viewed from above the loop.?</h3>
Given, $B(t)=(1.4 T) e^{-0.057 t}$




(b) 

c)
In conclusion, the direction of the induced current will be Counterclockwise.
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