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The square loop shown in the figure moves into a 0.80Tmagnetic field at a constant speed...

Question

The square loop shown in the figure moves into a 0.80Tmagnetic field at a constant speed...

The square loop shown in the figure moves into a 0.80Tmagnetic field at a constant speed of 10m/s. The loop has a resistance of 0.10?, and it enters the field at t = 0s.
The square loop shown in the figure moves into a 0

Science Physics
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Answers

10 cm

A(t)=2(x^2/2)=x^2=(vt)^2=v^2t^2

dA/dt=2v^2t

\varepsilon =-\frac{d\Phi }{dt}=-B\frac{\mathrm{d}A }{\mathrm{d} t}=-2v^2tB

\varepsilon =160t

because B=-0.8 T (into page) and v=10 m/s

I=V/R (ohm's law)

I=\frac{\varepsilon}{R} =\frac{160t}{0.1}=1600t

When 1/2 way:

10t=\frac{10\sqrt{2}}{200}\rightarrow t=0.00707\, s

at this t=

I=\frac{\varepsilon}{R} =\frac{160t}{0.1}=1600\cdot 0.00707=11.31\, A

So the current stops at 2t=0.01414 s

15 1(A) 10 E 0 5.0 10.0 15.0 20.0 t(ms)

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