Question

A rectangular loop of wire with length a=2.2 cm, width b=0.80 cm,and resistance R=0.40m ohms is placed near an infinitely long wirecarrying current i=4.7 A. The loop is then moved away fromthe wire at a constant speed v=3.2 mm/s. When the center ofthe loop is at distance r = 1.5b, what are the magnitude of themagnetic flux through the loop and the current induced in theloop?

Answer 1

Answer:

magnetic flux ΦB = 0.450324 ×$10^{-7}$  weber

current I = 1.02484 $10^{-8}$  A

Explanation:

Given data

length a = 2.2 cm = 0.022 m

width b = 0.80 cm = 0.008 m

Resistance R = 0.40 ohms

current I = 4.7 A

speed v = 3.2 mm/s = 0.0032 m/s

distance r = 1.5 b = 1.5 (0.008) = 0.012

to find out

magnitude of magnetic flux and the current induced

solution

we will find magnitude of magnetic flux thorough this formula that is

ΦB = ( μ I(a) /2 π ) ln [(r + b/2 ) /( r -b/2)]

here μ is 4π ×$10^{-7}$ put all value

ΦB = (4π ×$10^{-7}$  4.7 (0.022) /2 π ) ln [(0.012+ 0.008/2 ) /( 0.012 -0.008/2)]

ΦB = 0.450324 ×$10^{-7}$  weber

and

current induced is

current =  ε / R

current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]

put all value

current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]

current = 4π ×$10^{-7}$ (4.7) (0.022) (0.008) (0.0032) /  2π(0.40) [(0.012² ) - (0.008/2 )² ]

current = 1.02484 $10^{-8}$  A