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Ivanshal [37]
3 years ago
12

A fixed 15.3-cm-diameter wire coil is perpendicular to a magnetic field 0.77 T pointing up. In 0.20 s , the field is changed to

0.26 T pointing down. What is the average induced emf in the coil?
Physics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

Induced emf will be 0.468 volt

Explanation:

We have given diameter of wire d = 15.3 cm

So radius r=\frac{d}{2}=\frac{15.3}{2}=7.65cm

So area A=\pi r^2=3.14\times (7.65)^2=183.76\times 10^{-4}m^2

Change in magnetic field dB = 0.26 - 0.77 = -0.51 T

Time for change in magnetic field dt = 0.26 sec

We know that emf is given by e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-183.76\times 10^{-4}\times \frac{-0.51}{0.2}=0.0468volt

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The next four questions refer to the situation below.
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Answer:

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This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

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The subscripts are s for the swimmer, r for the river and g for the Earth

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now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

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with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

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This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

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