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lara [203]
3 years ago
11

A magnetic field is passing through a loop of wire whose area is 0.020 m2. The direction of the magnetic field is parallel to th

e normal to the loop, and the magnitude of the field is increasing at the rate of 0.24 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 2.1 T if the induced emf is to be zero? Use a minus sign if the area is to be shrunk
Physics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

a) E = 0.0048 Volts

b)  dA/dt = - 0.002285 m²/s

Explanation:

Given:

Area, A = 0.020 m²

Rate of change of magnetic field, dB/dt = 0.24 T/s

a) The magnitude of the emf induced (E) is given as:

  E= A × (dB/dt)

  on substituting the values in the above equation, we get

E = (0.020 m²) × (0.24 T/s)

or

E = 0.0048 Volts

b) Now, The induced emf when both the area and the magnetic field is varying

we have

E = B(dA/dt) + A(dB/dt)

Now, for the given case induced emf is zero i.e E = 0 and magnetic field B = 2.1 T

thus,

0 = (2.1 T)(dA/dt) + (0.020 m2)(0.24 T/s)

dA/dt = - 0.002285 m²/s

Hence, the area should be decreased at the rate of 0.002285 m²/s

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The mass of the planet Gallifrey is 8 times the mass of the Earth.

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  • let the radius of the Earth = R
  • gravitational field of Gallifrey = 2g
  • radius of Gallifrey = 2R

<h3>What is gravitational potential energy?</h3>
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The gravitational field strength of the Earth is given as follows;

g = \frac{GM}{r^2} \\\\G = \frac{gr^2}{M}

The gravitational field strength of the Planet Gallifrey is calculated as follows;

g_2 = \frac{GM_2}{r_2^2}

G = \frac{g_2r_2^2}{M_2} \\\\\frac{g_2r_2^2}{M_2} = \frac{gr^2}{M}\\\\M_2gr^2 = Mg_2r_2^2\\\\M_2 = \frac{Mg_2r_2^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times (2r)^2}{gr^2} \\\\M_2 = \frac{M \times 2g \times 4r^2}{gr^2} \\\\M_2 = 8M

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Learn more about gravitational field strength here:  brainly.com/question/14080810

4 0
2 years ago
The circumference of the earth is 40,000 km. The distance from the
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Answer:

The distance from the North Pole to the equator is 1*10^{7}m.

Explanation:

Circumference of Earth = 40,000 km    ......................(1)

Distance from the North Pole to the Equator is =  1/4th of the Circumference of Earth  ......................  (2)

Let Distance from the North Pole to the Equator be  d ,

the equation formed will be ,

d = 1/4 * Circumference of Earth       ........(3)......... ( from equation 1 )

put the value of Circumference of Earth in equation (3),

d = 1/4 * 40,000  km

d = 10,000 km

converting km to m ,

d = 10,000 * 10^{3} m

d =  1 * 10^{7} m

The distance from the North Pole to the equator is 1*10^{7}m.

7 0
3 years ago
A puck moves 2.35 m/s in a -22 degree direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50
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Answer:

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Explanation:

initial velocity of the puck is 2.35 m/s at -22 degree

so here it is given as

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v_i = 2.18 \hat i - 0.88 \hat j

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so we have

v_f = 6.42 cos50 \hat i + 6.42 sin50\hat j

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now displacement in Y direction is given as

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Answer:

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Hence, this is the required solution.

4 0
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