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Elena-2011 [213]

3 years ago

11

An elastic conducting material is stretched into a circular loop of 9.65 cm radius. It is placed with its plane perpendicular to

a uniform 1.14 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s. What emf is induced in volts in the loop at that instant?

1 answer:

Nadya [2.5K]3 years ago

8 0

Answer:

The induced emf in the coil is 0.522 volts.

Explanation:

Given that,

Radius of the circular loop, r = 9.65 cm

It is placed with its plane perpendicular to a uniform 1.14 T magnetic field.

The radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s , $\dfrac{dr}{dt}=-0.756\ m/s$

Due to the shrinking of radius of the loop, an emf induced in it. It is given by :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=B\dfrac{-d(\pi r^2)}{dt}\\\\\epsilon=2\pi rB\dfrac{dr}{dt}\\\\\epsilon=2\pi \times 9.65\times 10^{-2}\times 1.14\times 0.756\\\\\epsilon=0.522\ V$

So, the induced emf in the coil is 0.522 volts.

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From the Question we are told that

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Diano4ka-milaya [45]

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B. evaluation

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3 years ago

A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is

Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

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Solution,

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A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

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$\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}$

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The total potential is

$V_t=50912\ V+50912\ V=1\times 10^5\ V$

$\boxed{V_t=1\times 10^5\ V}$

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3 years ago