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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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Answer:

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Part C: 21.5%

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The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}

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Explanation:

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Here I will take refractive index of air and water

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Now let's look at the diagram I have attached here

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