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A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dAdt=−c, with c>0.

Answer :

Poltergeist

QUESTION:

Part A

The induced emf in the loop is measured to be [tex]V[/tex]. What is the magnitude [tex]B[/tex] of the magnetic field that the loop was in?

Part B

For the case of a square loop of side length [tex]L[/tex] being pulled out of the magnetic field with constant speed [tex]v[/tex] (see the figure), what is the rate of change of area [tex]c = -\dfrac{dA}{dt}[/tex]?

Answer:

Part A: [tex]B = -\dfrac{V}{c}[/tex]

Part B: [tex]c=-Lv[/tex]

Explanation:

Part A:

Faraday's law says that the induced voltage is equal to

[tex]V =-N \dfrac{d\Phi_B}{dt}[/tex],

which in our case(because we have only one loop) becomes

[tex]V =- \dfrac{d (BA)}{dt}[/tex],

and since the magnetic field is uniform (not changing),

[tex]V =-B \dfrac{dA}{dt}.[/tex]

Now, we know that [tex]\dfrac{dA}{dt} =c;[/tex]

therefore,

[tex]V =-B c[/tex]

which gives us

[tex]\boxed{B = -\dfrac{V}{c} }[/tex]

Part B:

The area of the loop can be written as

[tex]A = Lx[/tex],

where [tex]x[/tex] is the instantaneous length of the side along which the loop is moving.

Taking the derivative of both sides we get:

[tex]\dfrac{dA}{dt} = -L\dfrac{dx}{dt}[/tex],

and since [tex]v =\dfrac{dx}{dt}[/tex] we have

[tex]c = \dfrac{dA}{dt} = -Lv[/tex]

[tex]\boxed{c=-Lv}[/tex]

where the negative sign indicates that the area is decreasing.

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