when will I know that t will not be in the expression anymore?
The only way I can think it's possible is when $\theta=0$?
t vanishes in the expression when that is the case.
when I plugged in zero in the expression $\phi=BA$ now t vanished from the expression. taking its derivative would be...
This is what I am thinking when I say that emf should be zero when $\theta=90$,
We have $\phi = BA\cos(90°) = 0$ the flux is zero at $\theta=90°$
Now emf =$-N\frac{d\phi}{dt}$ but $\phi=0°$ so $\frac{d\phi}{dt}=0$
Now I have emf= $-N(0) = 0$. Tell me what I'am missing. I am thinking that...
Yes, thank you!
But there's one more thing I want to clarify.
I understand now that when the area vector is parallel to the magnetic field the flux is maximum becsuse the derivative at maximum or minimum of cos function is zero. But when the area vector is perpendicular to the field i.e...
hello guys!
I am confused about determining the equation for the induced emf in a rectangular coil with n turns rotating in a uniform magnetic field.
According to faraday's law of EMI the emf induced in a coil of wire is the rate of change of flux passing through it
$E_{induced}...
I just want to know how do I determine the number of terms will be in this summation. The answer to this 2N+1 terms. I can only arrive at preliminary steps of solving this. can you tell why 2N+1 is the number of terms? I know that the magnitude of complex exponential function squared would...
solve for theta
$\frac{\tan(\theta)(\sqrt{5-4\cos(\theta)}-1)}{\sqrt{5-4\cos(\theta)}}=\frac{10}{60}$
I have already tried my best solving this eqn but still couldn't get it. FYI getting that equation already took me a lot of work. Now I'am on the last piece of the problem I am solving which...
$240\sin(30^{\circ})+275\sin(40.9)=300$-->>$300=300$ $59.86\sin(30^{\circ})+275\sin(79.13)=300$---->>$300=300$
Does this mean that the values of W and $\theta$ that I get are valid solutions?
Actually the system of equations that I posted above came from a problem about equilibrium of a...
HI I LIKE YOU!:o
I just did what said and came up with a quadratic equation
$W^2-600W\sin(30)+300^2-275^2=0$
the two solutions are
$W=240.1$ and $W=59.86$
$\theta = 40.9$ and $\theta = 79.13$
which of them should I choose?
Can you help me how to solve this system of trig eqns
$W\cos(30^{\circ})-275\cos(\theta)=0$
$W\sin(30^{\circ})+275\sin(\theta)=300$
I have tried to divide the first eqn by 2nd and I get
$\tan(30^{\circ})=\frac{275}{300\cos(\theta)}-\tan(\theta)$ I'm stuck here! Kindly help me please!
finishing up with the problem,$\boldsymbol\tau=B_\circ I r^2 \boldsymbol{\hat \jmath}\int \cos^2 \phi\, d\phi$
$\boldsymbol\tau=B_\circ I r^2 \boldsymbol{\hat \jmath}\int_{0}^{\pi} \frac{1+\cos(2\phi)}{2}d\phi$
$\boldsymbol\tau=B_\circ I r^2 \boldsymbol{\hat \jmath}\int_{0}^{\pi}...
Now it's clear to me where those pieces came from!
What confused me is when you use integration in finding the torque, but that is also clear to me now.
Moving on, is $d\mathbf F_{lorentz}$ only for the right part of the curve or for the entire curve?
and also did you just change the cross...