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Nataly [62]
3 years ago
15

A circular coil that has 100 turns and a radius of 10.0 cm lies in a magnetic field that has a magnitude of 0.0650 T directed pe

rpendicular to the coil. (a) What is the magnetic flux through the coil? (b) The magnetic field through the coil is increased steadily to 0.100 T over a time interval of 0.500 s. What is the magnitude of the emf induced in the coil during the time interval?
Physics
1 answer:
yulyashka [42]3 years ago
6 0

Answer:

(a) 0.204 Weber

(b) 0.22 Volt

Explanation:

N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.

(a) Magnetic flux, Ф = N x B x A

Ф = 100 x 0.0650 x 3.14 x 0.1 0.1

Ф = 0.204 Weber

(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s

dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s

induced emf, e = N dФ/dt

e = N x A x dB/dt

e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V

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From the question, it will be travelling in North east direction because the the negative X components in vector is in horizontal direction which is the Eastern direction when using the four cardinal points and the Y components, the positive is in North direction if you use the four cardinal points. Therefore, the vector with both X components and Y components will be travelling in Northeast direction.

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Pls help me solve these questions.​
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The value of constant in terms of Force, k, A, V and ρ is  x = ln (F /kAρV ).

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The force which tends to oppose motion of fluid in forward direction like shear stress in liquids and friction force in solids

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A 5 kg toy is tied to a rope where the tension measures 150 N. What is the weight of the object?
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If you drop a rock with a density of 1.73 grams will it float or sink
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A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge o
aleksandrvk [35]

Answer:

a)  W = - 6.825 J,  b) θ = 1.72 revolution

Explanation:

a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

         W = ΔK

         W = K_f - K₀

          W = ½ m v_f² - ½ m v₀²

         W = ½ 0.325 (5.5² - 8.5²)

         W = - 6.825 J

b) find us the coefficient of friction

Let's use Newton's second law

            fr = μ N

y-axis (vertical)   N-W = 0

            fr = μ W

work is defined by

             W = F d

the distance traveled in a revolution is

             d₀ = 2π r

             W = μ mg d₀ = -6.825

            μ = \frac{ -6.825}{d_o \ mg}

               

The total work as the object stops the final velocity is zero v_f = 0

         W = 0 - ½ m v₀²

          W = - ½ 0.325 8.5²

          W = - 11.74 J

           μ mg d = -11.74

           

we subtitle the friction coefficient value

           ( \frac{-6.8525 }{d_o mg}) m g d = -11.74

               6.825  \frac{d}{d_o} = 11.74

               d = 11.74/6.825  d₀

               d = 1.7201  2π 0.400

               d = 4.32 m

this is the total distance traveled, the distance and the angle are related

              θ = d / r

              θ = 4.32 / 0.40

              θ = 10.808 rad

we reduce to revolutions

              θ = 10.808 rad (1rev / 2π rad)

              θ = 1.72 revolution

3 0
3 years ago
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