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andriy [413]
3 years ago
8

Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.012 m,

and carries a current of 6.2 A. The outer coil contains 220 turns and has a radius of 0.020 m. What must be the magnitude of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero
Physics
1 answer:
GREYUIT [131]3 years ago
5 0

Answer:

The current of the outer coil is  I_o   =   3.99 \ A

Explanation:

From the question we are told that

    The number of turns of the inner coil is  N_i  = 170 \ turn

    The radius of the  inner coil is R_i =  0.012 \ m

     The current of the inner coil is  I_i  =  6.2 \ A

      The number of turns of the outer coil is N_o  =  220 \ turns

      The radius of the  outer coil is R_o  =  0.02 0 \ m

       

Generally the net magnetic field is mathematically represented as

              B  =  \frac{N \ mu I }{ 2 * R  }

Now from told that the net magnetic field is common

So  

           \frac{N_i  \mu I_i} {2 * R _i} =  \frac{N_o  \mu I_o} {2 * R _o}

Here  \mu is the permeability of free space

making  I_o the subject

            I_o   =   \frac{ N_i  I_i *2 * R _o}{N_o  *2 * R _i}

substituting values

           I_o   =   \frac{ 170 *6.2 *2 * 0.020}{220   *2 * 0.012}

         I_o   =   3.99 \ A

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In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
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Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

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After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

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kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

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