A wire that is 0.20 meters long is moved perpendicularly through a magnetic field of strength 0.45 newtons/amperes·meter at a speed of 10.0 meters/second. What is the emf produced?

Question

shikamaru11 · Answer

so first configure the data given........... l=0.20m B=0.45N/Am V=10m/s and we've to find Emf=???

shikamaru11 · Answer

Emf = v * l * B, when v and B are perpendicular to each other..... so sin$$\sin \theta=1$$

shikamaru11 · Answer

hence put your data and here you go for the answer :)

michele_laino · Answer

the situation, can be represented by this drawing:
|dw:1452358525208:dw|
after an infinitesimal time $dt$, the wire has described the dashed rectangle, whose area is:
$dA=vdtL$, so if we apply the law of magnetic induction we get the requested $emf$, as below:
$$E = \frac{{d\phi }}{{dt}} = \frac{{BdA}}{{dt}} = \frac{{B \cdot v \cdot dt \cdot L}}{{dt}} = BvL$$
where, $d\phi$ is the infinitesimal change of magnetic flux of the field $B$, and $L$ is the length of the wire, furthermore $v$ is its speed

anonymous · Answer

I don't understand ow to do any of the, that's why I was just asking for the answer.

michele_laino · Answer

the requested value of the $emf$, is:
$$E = Blv = 0.45 \cdot 0.20 \cdot 10 = ...volts$$

anonymous · Answer

.9?

michele_laino · Answer

that's right!
we have an $emf$ of $0.9$ volts