Question

find an expression for the power expended pulling a conducting loopp out of a magnetic field.​

Answer 1

The equation would be $P= b^{2} v^{2} l^{2} /R$

Ohm Law:

Ohm’s law states the connection 'tween energetic current and potential dissimilarity. The current that flows through most leaders is straightforwardly equivalent to the service used to it.

Georg Simon Ohm, a German researcher was the first to confirm Ohm’s standard tentatively.Ohms society replies that the current running through the leader is straightforwardly equivalent to the potential distinctness across allure limits because the hotness and additional material environments are nonstop.

Let's derive the equation:

Consider a loop abcd moving with constant velocity v in a constant magnetic field.
a current is induced in the lopp in a clockwise durectuom and the loop segments being still in a magnetic field.

As the loop is moved to the right, teh area lying within the magnetic field increases,
P = (NLv/R).R
P= $b^{2} v^{2} l^{2} /R$

Hence we get our equation as $P= b^{2} v^{2} l^{2} /R$

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Answer 2

Given:

A conducting loop is being pulled out of a magnetic field

To find:

The power expended on performing the experiment

Solution:

Let the magnitude of the intensity of the magnitude field be B, the magnitude of the velocity with which the loop is being pulled out of the magnetic field be v and the length of the loop experiencing a deflecting force be L.

According to Lenz's law, to oppose the decrease in the magnetic flux, a clockwise current will be induced in the conducting loop.

Now, by Faraday's law, the magnitude of induced emf will be given by

E=BLv

And the magnitude of the induced current will be given by

I=BLv/R .....(Equation 1)

∴ The magnitude of deflecting force experienced by the loop,

F=iLB

Using equation 1, we get

F=(BLv/R)LB

⇒F=B²L²v/R .....(Equation 2)

∴ The power expended in pulling the loop out of the magnetic field,

P=Fv

Using equation 2, we get

P=(B²L²v/R)v

⇒P=B²L²v²/R

Therefore, the expression for the power expended in pulling a conducting loop out of a magnetic field is P=B²L²v²/R.

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