3. Refer to the diagram here. A steady, uniform magnetic field of magnitude B0 e

Question

3. Refer to the diagram here. A steady, uniform magnetic field of magnitude B0 exists in the shaded region. This field is directed perpendicularly into the page. You have a rectangular loop of rigid, conductive wire, of length L, width W, and resistance R, which you are pushing into the field at a steady, velocity v, as shown here:

The list of known values: B0, L, W, R, v, d, b, µ0.

Calculate the following, expressing your results using only known values.

a. While you are pushing the loop into the field, what force (magnitude and direction) does side 1 of the loop exert   on side 2? Why doesn’t the loop get deflected as a result of this force?

b. What minimum work must you do in order to push the wire loop a known distance d (where d < L) into the field?

c. Now let t = 0 at the moment when the loop is the distance d into the field. Suppose at that moment, the field   strength begins to vary according to this time function: B(t) = B0e–bt, where b is a known constant (with units of time–1). How much of the loop’s area will still be outside the field when the magnetic field in that external area   reverses direction? (You may assume that the reversal occurs while the loop’s right end is still in the field and its   left end is still outside the field.)

3. Refer to the di directed perpendicularly into the page. You have a rectangular loop of rigid, conductive wire, of length L, width W, and resistance R, which you are pushing into the field at a steady, velocity v, as shown here iagram here. A steady, uniform magnetic field of magnitude B, exists in the shaded region. This field is side 1 (loop velocity, v) side 2 The list of known values: Bo. L, W. R, v, d, b, Ho alculate the following, expressing your results using only known values a. While you are pushing the loop into the field, what force (magnitude and direction) does side 1 of the loop exert on side 2? Why doesn't the loop get deflected as a result of this force? What minimum work must you do in order to push the wire loop a known distance d (where d

Explanation / Answer

We can solve this problem using the Faraday induction law

E = - d E / dt

E = B A Cos

We can see that an electromotive force is induced in the perpendicular face of the loop (W), we can also see that there is a magnetic force F = q v x moving charge carriers electrons down creating a current in the opposite direction, cleans the current it goes from side 2 to side 1

E = B ( x L)

E = - B W dx/dt

E = - B W v

E = R I

I = E /R

I = - B w v /R (1)

This is the induced current in the loop

as we can see the current flows in the 1 to the left side and on the side 2 to the right

We can calculate the force between these two wire where a current flows, the strength of side 1 by the magnetic field created by the current of side 2

F1 = I1 L B2

B2 = o(I2) / 2 w

I1 = -I2 even if the opposite direction

F1 =- o I I L / (2 w)

F1 = - o I2 L / (2 w)

This force is exerted outwardly on each wire and is counteracted by the deforming force of young that inwardly directed

Part b)

W = F.d

F = I B w

W = I B w x

Replace 1

W= ( B w v /R ) B w x

W = B2w2 v x /R

Part c)

We calculate the time for the field reaches zero

B = Bo e(-b t)

B/Bo = e -b t

Bo/B = e b t

ln (Bo/B ) = B t

t = 1/b ln (Bo/B)

we can see that B = 0 implies infinite time

We calculate a value for small magnetic field B= Bo/ 10000

t = 1/b ln (104)

t = 9.2/b

the loop moves with a constant speed, so we use expressions kinematic

v = (X-Xo) /t

X = Xo + v t

X -Xo = v t

X = v (1/b) ln (Bo/B)

This is the distance that the loop progresses before the change of direction of the field,

for the explicit case

X = (v/b) 9.2

What is left out is

Xout = L - d – X

Xout = L - d – (v/b) 9.2

this is what the loop off the field when a decreased factor 104

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