Question

A rectangular loop of wire with width w = 2.50 cm and height h 5.00 cm is placed as pictured here a distance a- 1.50 cm from

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Answer #1

Given

w -2,50cm0,025m

h 5.00cm-0.05m

k = 354.4/s2

I=kt^{2}

a 1.5cm= 0.015m

The magnetic field produced by the infinite wire at r distance is given by

B=\frac{\mu_{o}}{4\pi}\frac{I}{r}

The flux through the rectangle is given

\Phi=\int_{a}^{a+w} Bhdr=\int_{a}^{a+w}\frac{\mu_{o}}{4\pi}\frac{I}{r}h dr

or,

4π 0

or,

4π 0

The EMF of the wire is given by

dt

or,

\epsilon =-2\frac{\mu_{o}}{4\pi}kth\times ln\frac{a+w}{a}=-\frac{\mu_{o}}{2\pi}kth\times ln\frac{a+w}{a}

The emf in the wire at t=3.00s is given by

\epsilon =-\frac{4\pi \times 10^{-7}}{2\pi}\times 354\times 3\times 0.05\times ln\frac{0.015+0.025}{0.015}

or,

є =- 10.4 × 10-бу

or,

10.4M (Counter-clockwise)

The EMF is at t=3s is 10.4\mu V counter-clockwise.

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Answer #2

i-kt tuo 27 at 27 20 . 8 メ人V j in anti-clockwise

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