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Makovka662 [10]
3 years ago
14

A 200 turn coil is in a uniform magnetic field that is decreasing at the rate 0.20 T/s. The coil is perpendicular to the field a

nd its dimensions are 0.20 m by 0.40 m. What is the magnitude of the induced emf in the coil
Physics
1 answer:
sattari [20]3 years ago
8 0

Answer:

emf = 3.2V

Explanation:

In order to calculate the magnitude of the induced emf in the coil you use the following formula:

emf=-N\frac{d\Phi_B}{dt}       (1)

N: turns of the coil = 200

ФB: magnetic flux = A*B

A: area of the coil = (0.20m)(0.40m) = 0.08m²

B: magnitude of the magnetic field

You take into account that the area of the coil is constant, while magnetic field changes on time. Then, the equation (1) becomes:

emf=-NA\frac{dB}{dt}         (2)

dB/dt =  rate of change of the magnetic field = -0.20T/s (it is decreasing)

You replace the values of all parameters in the equation (2):

emf=-200(0.08m^2)(-0.20T/s)=3.2V

The induced emf in the coil is 3.2V

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Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
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Answer:

A. Zero

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                                                                                                        = 0.3 m

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The energy stored in the charge at 0.3 m is given by the formula,

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The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

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The energy stored in it is,

                             W = V₂ - V₁

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Hence, the energy stored in the charge is, W = 0        

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