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V125BC [204]
3 years ago
15

An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno

id's edge, however, but extends into the area around the magnet. Suppose a technician walks toward the scanner at 0.80 m/s from a region 1.0 m from the scanner where the magnetic field is negligible, into a region next to the scanner where the field is 6.0 T and points horizontally. As a result of this motion, what is the maximum magnitude of the change in flux through a loop defined by the outside of the technician's head? Assume the loop is vertical and has a circular cross section with a diameter of 19 cm.
What is the magnitude of the average induced emf around the outside of the technician's head during the time she's moving toward the scanner?
Physics
1 answer:
GaryK [48]3 years ago
7 0

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

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Answer:

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Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

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v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

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2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

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\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

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2(V_1+9.0)^2=(2V_1+9.0)^2

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\sqrt{2}(V_1+9.0)=(2V_1+9.0)

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V_2=2\times 8.98m/s = 17.96m/s

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Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

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