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dolphi86 [110]
3 years ago
5

A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current w

ill be greatest when the magnet is located so that the loop is
Physics
2 answers:
Ksivusya [100]3 years ago
8 0
<span>A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current will be greatest when the magnet is located so that </span>the loop is near either the north or the south pole.
noname [10]3 years ago
3 0

Answer:

Loop must be near the end of the pole of bar magnet

Explanation:

As we know by Faraday's law of electromagnetic induction the induced EMF or induced current in the loop is given by

EMF = \frac{d\phi}{dt}

here we know that

\phi = B.A

now we know that as the magnet comes closer to the loop the magnetic field due to any of the pole of magnet will increase.

So here the maximum flux will pass through the magnet when magnet is closer to the loop.

So here when A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current will be greatest when the magnet is located so that the loop is when Loop must be near the end of the pole of bar magnet

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A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is co
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Answer:

A) μ = A.m²

B) z = 0.46m

Explanation:

A) Magnetic dipole moment of a coil is given by; μ = NIA

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We are given

N = 300 turns; I = 4A ; d =5cm = 0.05m

Area = πd²/4 = π(0.05)²/4 = 0.001963

So,

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B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;

B = (μ_o•μ)/(2π•z³)

Let's make z the subject ;

z = [(μ_o•μ)/(2π•B)] ^(⅓)

Where u_o is vacuum permiability with a value of 4π x 10^(-7) H

Also, B = 5 mT = 5 x 10^(-6) T

Thus,

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3 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
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Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

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The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

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