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2 years ago

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Suppose the first coil of wire illustrated in the simulation is wound around the iron core 7 times. Suppose the second coil is w

ound around the core N2 = 13 times. Shortly after the switch has been closed, the flux through the second coil of wire changes from 2.1 Wb per turn to 2.7 Wb in each turn of wire over a time interval of 0.3 s. Find the magnitude of the emf induced in the coil.

2 answers:

Elena L [17]2 years ago

8 0

Answer:

$emf=26\ V$

Explanation:

Given:

no. of turns in the first coil, $n_1=7$

no. of turns in the second coil, $n_2=13$

the change in flux through the secondary coil, $d\phi=2.7-2.1=0.6\ Wb$

time taken for the change in flux, $dt=0.3\ s$

According to the Faraday's law there will be an emf induced in the coil associated with the rate of change in magnetic flux.

<u>This emf is mathematically given as:</u>

$emf=n_2\times \frac{d\phi}{dt}$

$emf=13\times \frac{0.6}{0.3}$ (we take no. of turns of the second coil because the rate of change in flux is associated with the second coil)

$emf=26\ V$

SVETLANKA909090 [29]2 years ago

5 0

Answer:

Emf induced in the coli will be equal to 26 volt

Explanation:

We have given number of turns N = 13

It is given that magnetic flux changes from 2.1 Weber to 2.7 Weber in 0.3 sec

Change in flux $d\Phi =2.7-2.1=0.6Weber$

Change in time dt = 0.3 sec

We have to find the induced emf

Induced emf is equal to $e=N\frac{d\Phi }{dt}=13\times \frac{0.6}{0.3}=26volt$

So emf induced in the coil will be equal to 6 volt

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1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

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1)

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$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

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On the other hand, the potential at the surface is

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The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

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- Then at r = R, the potential is $V(R)$

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