A conducting loops of wire has an area of 250 cm^2 and a resistance of 40 ohm. There is a magnetic field of 200T perpendicular to the loop. At what rate must this field be reduced to induce a current o f.2A in the loop?

Question

nali · Answer

your could use the equation 

$$\large V= - N \Delta (\phi B)/\Delta (t) = -N (B A \cos \theta)/ t = - N A \cos (\theta) (\Delta B/\Delta(t))  $$

nali · Answer

First find the voltage using Voltage = Current times Resistance 
or V= I times R 

Then plug in the V and find the ratio of B/t (magnetic field over time)

nali · Answer

@Grazes

anonymous · Answer

I have v=8V. Now what?

nali · Answer

do you mean 80 V ?

anonymous · Answer

V=IR=(.2A)(40ohm)=8V

nali · Answer

oh sorry did not notice the point in front of the 2 there

nali · Answer

anyway, then you plug in V and the area after changing it to m and the magnetic field

anonymous · Answer

8V=−N(2.5m^2)(ΔB/Δ(t))

Now what?

nali · Answer

|dw:1398656360718:dw|

nali · Answer

cos 90 = 1 so what you have is right so far

anonymous · Answer

cos90=0....
perpendicular means cos0

anonymous · Answer

But I got it, thanks.

nali · Answer

oh yeah sorry again but anyway that is good you got what I did

nali · Answer

by the way, the question is asking for the ratio, so after finding the time you still need to divide magnetic field by the time to get the ratio

anonymous · Answer

the ratio is just B/t, isn't it?

anonymous · Answer

It's one term, no need to divide.

nali · Answer

no because you do not have the time (exact amount of seconds) so you need to find it first and then you could find the ratio