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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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A 900 N student runs up the stairs 3.5 m high in 12<br> seconds. How much POWER do they generate?
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4 0
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Vector a has a magnitude of 12.3 units and points due west. vector b points due north. what is the magnitude of b if a - b has a
algol [13]

The magnitude of vector b is 8.58 Unit.

Since both the vectors a and b are perpendicular to each other, so we can apply the Pythagoras theorem to calculate the magnitude of the vector b.

Applying the Pythagoras theorem

(a-b)^2=a^2+b^2

15^2=12.3^2-b^2

b=8.58 unit

Therefor the magnitude of the vector b is 8.58 unit.

8 0
3 years ago
Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various
Alexxx [7]

Answer:

Q_{2}=1200cm^{3}/s

Explanation:

Given data

Q₁=200cm³/s

We know that:

F=n\frac{vA}{l}\\

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s

3 0
2 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
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