Question

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.

Answer 1

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)