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I am Lyosha [343]
3 years ago
9

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

= 0.285 square m.Suppose the radius of the elastic loop increases at a constant rate, dr/dt = 3.70 cm/s .
A) Determine the emf induced in the loop at t = 0.
B)Determine the emf induced in the loop at t = 1.00 s .
Physics
1 answer:
love history [14]3 years ago
5 0

Answer:

Part a)

EMF = 14 \times 10^{-3} V

Part b)

EMF = 15.67 \times 10^{-3} V

Explanation:

As we know that magnetic flux through the loop is given as

\phi = B.A

now we have

\phi = B\pi r^2

now rate of change in flux is given as

\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}

now we know that

A = \pi r^2

0.285 = \pi r^2

r = 0.30 m

Now plug in all data

EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)

EMF = 14 \times 10^{-3} V

Part b)

Now the radius of the loop after t = 1 s

r_1 = r_0 + \frac{dr}{dt}

r_1 = 0.30 + 0.037

r_1 = 0.337 m

Now plug in data in above equation

EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)

EMF = 15.67 \times 10^{-3} V

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Using Rydberg's Equation:

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Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

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E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

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3 years ago
Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
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7800kg/m³

Explanation:

Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is

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7.8g = 7.8/1000 kg

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Answer:

Ff = 839.05 N

Explanation:

We can use the equation:

Ff = μ*N

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