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Keith_Richards [23]
3 years ago
12

A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

f the entire wire is reshaped from a two turn circle to a one-turn circle in 0.0572 s (while remaining in the same plane), what is the magnitude of the average induced emf E in the wire during this time?
Physics
1 answer:
Basile [38]3 years ago
6 0

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

E=-\dfrac{d\phi}{dt}

E=-B\dfrac{dA}{dt}

E=-B\dfrac{A_{2}-A_{1}}{dt}

E=B\dfrac{A_{1}-A_{2}}{dt}.....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

l=2\times 2\pi\times r_{1}

l=2\times 2\pi\times 0.63

l=7.916\ m

The length when wire is in one loop

l=2\pi\times r_{2}

7.916=2\times \pi\times r_{2}

r_{2}=\dfrac{7.916}{2\times \pi}

r_{2}=1.259\ m

We need to calculate the initial area

A_{1}=N\times\pi\times r_{1}^2

Put the value into the formula

A_{1}=2\times3.14\times(0.63)^2

A_{1}=2.49\ m^2

The final area is

A_{2}=N\times\pi\times r_{2}^2

A_{2}=1\times\pi\times(1.259)^2

A_{2}=4.98\ m^2

Put the value of initial area and final area in the equation (I)

E=0.219\dfrac{2.49-4.98}{0.0572}

E=-9.533\ V

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

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lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

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De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

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a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

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b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

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The value is q = 3.4 *10^{-6} \ C

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