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seropon [69]
3 years ago
6

A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant

rate of 11.0cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf induced in the loop after exactly time 4.00s has passed since the circumference of the loop started to decrease AND find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Physics
1 answer:
Travka [436]3 years ago
4 0

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}\ \ =-B\frac{A_2-A_1}{t_2-t_1}   (1)

ФB: magnetic flux =  BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm

To find the areas A1 and A2 you calculate the radius:

r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2

Finally, the emf induced, by using the equation (1), is:

emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

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erik [133]

Answer:

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A 133 kg horizontal platform is a uniform disk of radius 1.95 m and can rotate about the vertical axis through its center. A 62.
cupoosta [38]

Answer:

The moment of inertia of the system is  I = 400.5 \ kg \cdot m^2

Explanation:

From the question we are told that

    The mass of the platform is  m =  133\ kg

     The  radius of the  platform is  r = 1.95 m

     The mass of the person is m_p  =  62.7 \ kg

     The position of the person from the center is  d =  1.19 \ m

       The mass of the dog is m_D  =  28.5 \ kg

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Generally the moment of inertia of the platform with respect to its axis is  mathematically represented as

       I_p  =  \frac{m r^2}{2}

The  moment of inertia of the person with respect to the axis is mathematically represented as

        I_z  =  m_p* d^2

The  moment of inertia of the dog with respect to the axis is mathematically represented as

       I_D =  m_d *  D^2

So the moment of inertia of the system about the axis  is mathematically evaluated as

        I  = I_p + I_z + I_D

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substituting values  

            I = \frac{(133) * (1.95)^2}{2}  +  (62.7) * (1.19)^2 +  (28.5) * (1.45)^2

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8 0
3 years ago
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Now, plug in the numbers and solve for acceleration:
a = \frac{335}{70} = 4.79

So, your answer is: 4.79 m/s².
5 0
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