A1.80 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?

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 (0.018) / 2π × ln((0.012+0.018)/0.012) × 100

d∅ /dt  = 3.3 × V

so

current = (d∅ /dt)  / R

current = 3.3 × / 0.0120

current = 27.5 × A

so current is 27.5 μA

the motion of a pendulum is a classic example of mechanical energy conservation. a pendulum consists of a mass (known as a bob ) attached by a string to a pivot point. as the pendulum moves it sweeps out a circular arc, moving back and forth in a periodic fashion.