A2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 ω has one edge parallel to a long straight wire. the near edge of the loop is 1.20 cm from the wire. the current in the wire is increasing at the rate of 120 a/s . what is the current in the loop?

current in loops is 52.73 μA

Explanation:

given data

side of square a = b  = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c  = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π××0.024 / 2π(1.20×) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 ×× 1.09861 × 120

current = 52.73 ×  A

so here current in loops is 52.73 μA

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v=ir = 3x15 =45 v