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natka813 [3]
3 years ago
7

Nolan set up the equipment shown in order to induce a current in a wire. Which best explains why no current is induced?

Physics
2 answers:
slava [35]3 years ago
3 0

Answer:

the answer is C on edge

Explanation:

liq [111]3 years ago
3 0

Answer:

(C)The magnet needs to be moved through the coils of wire.

Explanation:

As we know that rate of change in magnetic flux linked with a closed loop will induce the EMF in the closed loop

so it is given as

V = -\frac{d\phi}{dt}

here we know that

\phi = B.A

now here in order to change the flux we need to change the magnetic field linked with the closed loop

so here when we move the magnet then the magnetic field of the magnet linked with the coil will also change and that will induce the current in the coils

So correct answer is

(C)The magnet needs to be moved through the coils of wire.

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So we can conclude that electromagnetic waves like light do not require medium for its propagation.

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The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th
pickupchik [31]

Answer:

a

The focal length of the lens in water is  f_{water} = 262.68 cm

b

The focal length of the mirror in water is  f =79.0cm

Explanation:

From the question we are told that

    The index of refraction of the lens material = n_2

    The index of refraction of the medium surrounding the lens = n_1

 

The lens maker's formula is mathematically represented as

            \frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

Where f is the focal length

            n is the index of refraction

            R_1 and R_2 are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air  we have  

           \frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

    When immersed in liquid the formula becomes

          \frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2}  ]

The ratio of the focal length of the the two medium is mathematically evaluated as

           \frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }

From the question

      f_{air }= 79.0 cm

       n_2 = 1.55

and the refractive index of water(material surrounding the lens) has a constant value of  n_1 = 1.33

         \frac{f_{water}}{79}  = \frac{1.55- 1}{\frac{1.55}{1.44}  -1}

           f_{water} = 262.68 cm

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

   

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