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Morgarella [4.7K]
3 years ago
7

A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin

g instrument whose resistance is 45.5 O. When the coil is rotated quickly from a position parallel to a uniform magnetic field to one perpendicular to the field, the instrument indicates a charge of 3.53 x 10-5 C.What is the magnitude of the magnetic field?
Physics
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

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Answer:

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The speed of the plane is 1.2131147541 m/s

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K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J

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Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

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Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

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The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]  

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