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Oliga [24]
3 years ago
7

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

and resistivity of 2.18 × 10-8Ω·m. At what rate must the magnitude of the magnetic field change to induce a 11 A current in the loop?
Physics
1 answer:
I am Lyosha [343]3 years ago
8 0

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

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Answer:

    v₂ = 63.62 m / s

Explanation:

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We will assume that the distance between the two parts is small, so y₁ = y₂

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we substitute

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we substitute

         v₂² = \frac{1000}{1 \ 1.29 \ 9.8} + 63^2

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7 0
3 years ago
How much do you have to increase the vertical distance to double the impact speed of an object?
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A horizontal force of 30N is applied to a mass of 10 kg causing it to accelerate. If the coefficient of friction is 0.20 what is
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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.

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In this question ,

surface vertical force = Weight of the object

Thus ;

svf = ( mass ) × ( gravity acceleration )

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svf = 10 × 10 = 100 N

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If gravity acceleration is 9.8 :

svf = 10 × 9.8 = 98 N

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frictional force = 98 × 0.20

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Answer: Option (a) is the correct answer.

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