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Elenna [48]
3 years ago
5

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

ic field. A field of 1.7T is directed along the positive z-direction, which is upward.
Part A: If the loop is removed from the field region in a time interval of 2.4

ms , find the average emf that will be induced in the wire loop during the extraction process.



Part B:If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?



Could you please explain this in a step by step process, and include reasoning to every assumption in relation to the Right Hand Rule? Thanks.
Physics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

See answer

Explanation:

The area of the circular loop is given by:

A = \pi r^2

The magnetic flux is given by:

\phi = \int \vec{B} \cdot d\vec{A}

d\vec{A} is parallel to \vec{B} and \vec{B} is constant in magnitude and direction therefore:

\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2

Part A)

initially the flux is \phi =\pi B r^2

after the interval \Delta t= 2.4 [m/s]

the flux is

\phi = 0

now, the EMF is defined as:

\epsilon =- \frac{d \phi}{dt},

if we consider \Delta t= 2.4 [m/s] very small then we can re-write it as:

\epsilon =- \frac{\Delta \phi}{\Delta t}

\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12

then:

\epsilon =- \frac{-0.12}{0.0024} = 50 [V]

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

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Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge
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Answer:

Electric charge in the earth will be Q=5.231\times 10^5C

Explanation:

We have given that E = 116 N/C

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We have to find the electric charge in the earth '

We know that electric field due to charge is given by E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}. here K is coulomb's constant

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Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe
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Answer:

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<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

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We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = \pi r^2h

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

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h = 0.025/0.0531 = 0.47 m

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ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

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therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

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