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I am Lyosha [343]
4 years ago
6

A fixed coil of wire with 10 turns and an area of 0.055 m2 is placed in a perpendicular magnetic field. This field oscillates in

direction and magnitude at a frequency of 10 Hz and has a maximum value of 0.12 T, (a) What is the average emf induced in the coil during the time it takes for the field to go from its maximum value in one direction to its maximum value in the other direction? (b) Repeat part (a) for a time interval of one complete cycle. (c) At what time(s) during a complete magnetic field cycle would you expect the induced emf to have its maximum magnitude? What about its minimum value? Explain both answers
Physics
1 answer:
Blababa [14]4 years ago
8 0

Answer:

Part a)

Average EMF for half cycle is

E_{avg} = 2.64 V

Part b)

For one complete cycle we will have

E_{avg} = 0

Part c)

Maximum induced EMF will be at

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = 0.05s and 0.1 s

Explanation:

As we know that magnetic field is oscillating in direction as well as magnitude

so induced EMF is given as

E = NBA\omega sin(\omega t)

Part a)

For average value of EMF from positive maximum to negative maximum which is equal to half cycle

so we have

E_{avg} = NBA\omega \frac{2}{T}\int_0^{T/2} sin(\omega t) dt

E_{avg} = \frac{2NBA\omega}{\pi}

E_{avg} = \frac{2(10)(0.12)(0.055)(2\pi (10))}{\pi}

E_{avg} = 2.64 V

Part b)

For one complete cycle we will have

E_{avg} = NBA\omega \frac{1}{T}\int_0^T sin(\omega t) dt

E_{avg} = 0

Part c)

Maximum induced EMF will be at

t = \frac{T}{4} and \frac{3T}{4}

here we know

T = \frac{1}{f} = 0.1 s

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = \frac{T}{2} and T

so it is

t = 0.05s and 0.1 s

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Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to
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Answer:

Explanation:

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Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

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3 years ago
A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

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