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Ilia_Sergeevich [38]
3 years ago
12

An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (

in V) is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms? V †
Physics
1 answer:
lara31 [8.8K]3 years ago
3 0

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV

Therefore, the average emf induced in the coil is 175 mV

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Answer:

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Explanation:

The change in potential energy of the man is given by:

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where

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In this problem, we have:

\Delta U=4620 J is the gain in potential energy

g = 9.8 m/s^2 is the gravitational acceleration

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Answer:

2a.

a=1.13ms^-2

2b.

S=277m

2c.

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S=ut+1/2 at²

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2c.

V=U+AT

V=22+1.13(5)

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Answer:

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For electric field outside cylinder, check image 03 attached

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