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34kurt
4 years ago
10

A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the

axis of the loop. Somebody carefully grabs the edges of the loop and begins pulling them apart such that the area of the loop increases at a rate of 20 cm2/s. What is the magnitude of the induced EMF in the loop?
Physics
1 answer:
Fofino [41]4 years ago
7 0

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
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Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

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E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

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Using formula of electric field

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E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

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Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

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