A2.40 cm × 2.40 cm square loop of wire with resistance 1.30×10^−2 ω has one edge parallel to a long straight wire. the near edge of the loop is 1.10 cm from the wire. the current in the wire is increasing at the rate of 130 a/s . what is the current in the loop?

current in loops is 55.41 μA

Explanation:

given data

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

resistance R = 1.30×10^−2 Ω

edge of the loop c  = 1.10 cm = 0.011 m

rate of current = 130 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................1

so current = 1/R × d∅/dt

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

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........3

so from equation 1 we get here current

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

current = ( 4π× ×0.024 / 2π(1.30× ) × ln (0.024 + 0.011/0.011) × 130

solve it and we get

current = 3.6923 × × 1.15745 × 130

current = 55.41 × A

so current in loops is 55.41 μA

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