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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?

Respuesta :

Answer:

magnetic flux ΦB = 0.450324 ×[tex]10^{-7}[/tex]  weber

current I = 1.02484 [tex]10^{-8}[/tex]  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 ×[tex]10^{-7}[/tex] put all value

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

ΦB = 0.450324 ×[tex]10^{-7}[/tex]  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 ×[tex]10^{-7}[/tex] (4.7) (0.022) (0.008) (0.0032) /  2π(0.40) [(0.012² ) - (0.008/2 )² ]

current = 1.02484 [tex]10^{-8}[/tex]  A

onyebuchinnaji

The magnitude of the flux through the loop is [tex]1.433 \times 10^{-8} \ Wb[/tex].

The current induced in the loop is [tex]1.034 \times 10^{-5} \ A[/tex].

The given parameters;

  • length of the rectangular loop, a = 2.2 cm
  • width of the loop, b = 0.8 cm
  • resistance, R = 0.4m ohms
  • current in the wire, I = 4.7 A
  • the center of the loop, r = 1.5b
  • speed of the loop, v = 3.2 mm/s

The magnitude of the flux through the loop is calculated as follows;

[tex]\Phi _B = \int\limits {B} \, dA \\\\\Phi _B = \int\limits^{r + b/2}_{r - b/2} {(\frac{\mu_o I}{2\pi r}) } \, adr\\\\ \Phi _B = (\frac{\mu_o I a}{2\pi }) \ ln(r)\\\\ \Phi _B = (\frac{\mu_o I a}{2\pi }) \ ln (\frac{r \ +\ b/2}{r\ - \ b/2} )\\\\\Phi _B = (\frac{\mu_o I a}{2\pi }) \ ln (\frac{1.5b \ +\ b/2}{1.5 b\ - \ b/2} )\\\\\Phi _B = (\frac{\mu_o I a}{2\pi }) \ ln ( 2.0)\\\\\Phi _B = (\frac{4\pi \times 10^{-7} \times 4.7 \times 0.022 }{2\pi }) \ ln ( 2.0)\\\\|\Phi _B| = 1.433 \times 10^{-8} \ Wb[/tex]

The current induced in the loop is calculated by applying Faradays law;

[tex]\varepsilon = IR\\\\\frac{d\Phi_B }{dt} = IR\\\\I = \frac{d\Phi_B }{R dt} = (\frac{\Phi_B }{R} )[\frac{d}{dt} \ ln(\frac{r + b/2}{r - b/2})]\\\\I = \frac{\mu_0 I a}{2\pi R} (\frac{bv}{r^2\ - \ (b/2)^2} ) = \frac{\mu_0 I a}{2\pi R} (\frac{bv}{2b^2} ) \\\\I = \frac{4\pi \times 10^{-7} \times 4.7 \times 0.022}{2\pi \times 0.4 \times 10^{-3}} (\frac{0.008 \times 3.2 \times 10^{-3} }{2( 0.008)^2} )\\\\I = 1.034 \times 10^{-5} \ A\\\\[/tex]

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