Question

A 2.40 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?

Answer 1

Answer:

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× $10^{-7}$ ×0.024 / 2π(1.30× $10^{-2}$ ) × ln (0.024 + 0.011/0.011) × 130

solve it and we get

current = 3.6923 × $10^{-7}$ × 1.15745 × 130

current = 55.41 × $10^{-6}$ A

so current in loops is 55.41 μA