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3 years ago

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One hundred turns of insulated copper wire are wrapped into a circular coil of crosssectional area 1.20⇥103 m2. The two ends of

the wire are connected to an ammeter. The total resistance in the circuit is 1.30 ⌦. If an externally applied uniform magnetic ﬁeld perpendicular to the plane of the coil changes from 1.60 T in one direction to 1.60 T in the opposite direction in 1.5 seconds, what is the current through the ammeter? Hint: Start with Faraday’s Law.

1 answer:

arsen [322]3 years ago

5 0

Answer:

236.3 x $10^-^3$ C

Explanation:

Given:

B(0)=1.60T and B(t)=-1.60T

No. of turns 'N' =100

cross-sectional area 'A'= 1.2 x $10^-^3$ m²

Resistance 'R'= 1.3Ω

According to Faraday's law, the induced emf is given by,

ℰ=-NdΦ/dt

The current given by resistance and induced emf as

I = ℰ/R

I= -NdΦ/dtR

By converting the current to differential form(the time derivative of charge), we get

$\frac{dq}{dt}$ = -NdΦ/dtR

dq= -N dΦ/R

The change in the flux dФ =Ф(t)-Ф(0)

therefore, dq = $\frac{N}{R}$ (Ф(0)-Ф(t))

Also, flux is equal to the magnetic field multiplied with the area of the coil

dq = NA(B(0)-B(t))/R

dq= (100)(1.2 x $10^-^3$ )(1.6+1.6)/1.3

dq= 236.3 x $10^-^3$ C

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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

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$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

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(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

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$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

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According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

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