Question

A 1.80 cm × 1.80 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 100 A/s . What is the current in the loop?

Answer 1

Answer:

current in loop is 27.5 μA

Explanation:

given data

side a = b = 1.80 cm = 0.018 m

resistance 1.20×10^−2 Ω

edge distance = 1.20 cm = 0.012 m

to find out

current in loop

solution

we know here rate dI/dt = 100 A/s

so here current = 1 / R × d∅/dt     ..............a

and

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

differentiate it

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

now put all these value and find d∅ /dt

d∅ /dt  = 4π $10^{-7}$ (0.018) / 2π × ln((0.012+0.018)/0.012) × 100

d∅ /dt  = 3.3 × $10^{-7}$ V

so

current = (d∅ /dt)  / R

current = 3.3 × $10^{-7}$ / 0.0120

current = 27.5 × $10^{-6}$ A

so current is 27.5 μA