Recent content by Hamal_Arietis

Thank you, I did not concern about the wires.

a) the magnetic field does not change if the magnet rotates, but the conductor rotate so EMF is induced and I found the answer for a) situation In b) situation, I think it is so strange because there are no relative movement between conductor and magnetic field. And if you ask me what part of...

one connect to the axle and the other end to a sliding contact, you can see in the image.

I think Voltmeter's needle will not jump in both situations. But the answer is yes. I am going to use this to explain why. The reasons to create the voltage are by - The changing by time of Magnetic field - The move of the object in magnetic field In both situations, they are not occur so I...

I tried it. And I found the answer. by considering ##d=d_0## as the potential origin. The potential is $$V=\frac{\sigma}{\pi\epsilon_0}(a\log{(\frac{d_0^2+a^2}{d^2+a^2}})^\frac{1}{2}+d_0\tan^{-1}(\frac{a}{d_0})-d\tan^{-1}(\frac{a}{d}))$$ It seems correct because when I calculate the limit of V...

$$\int\frac{dx}{\sqrt{x^2+d^2}}=log |x+\sqrt{x^2+d^2}|+C$$ but my fomular is $$ \int\frac{rdr}{\sqrt{r^2+d^2}}=\sqrt{r^2+d^2} +C$$ the problem is when I replace ##r=\frac{a}{cos\theta}##, the integral became more complicated. the double integral became $$\int_0^{\frac{\pi}4}...

I found out the equation of electric potential, that is V=\int_{-a}^a \int_{-a}^{a} \frac{σdxdy}{4 \pi \epsilon_0\sqrt{x^2+y^2+d^2}}=\int_{0}^a \int_{0}^{a} \frac{σdxdy}{\pi \epsilon_0\sqrt{x^2+y^2+d^2}} but I couldn't calculate the integral. It seems convenient if we use the polar coordinate...

$$A=\int_0^{\frac{\pi}{a}}f(t)\cos at\,dt$$ $$A=\int_0^{\frac{\pi}{a}}( 1 + A \cos 2at + B \sin 2at)cos(at)dt$$ $$A=\frac{4B}{3a}$$ And $$B=\int_0^{\frac{\pi}{a}}( 1 + A \cos 2at + B \sin 2at)sin(at)dt$$ $$B=\frac{2}{a}+\frac{2A}{3a}=\frac{6+2A}{3a}$$ Then solve both equation...

Sorry, I don't understand your idea. Can you give me some document about it?. This problem is an examination for high school students to entrance university. I have learned that: If ##F(x)=\int_a^xf(t)dt## then: $$\frac{d}{dx}F(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$$

I try to find f'(x) by the fundamental of calculus before but it don't have answer

Both equations are: $$f(t)=A+1$$ $$f(t)=1+Acos(2at)+Bsin(2at)$$ So ##A=Acos(2at)+Bsin(2at)## and I transfer it into: $$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt$$

Oh, I see: $$A=Acos(2at)+Bsin(2at)$$ $$A(1-cos(2at))=Bsin(2at)$$ $$2Acos^2(at)=2Bsin(at)cos(at)$$ $$cos(at)(Acos(at)-Bsin(at))=0$$ So we have: $$\int_0^{\frac{\pi}{a}}f(t)cos^2(at)dt=\int_0^{\frac{\pi}{a}}f(t)sin^2(at)dt \Rightarrow f(t)cos^2(at)=f(t)sin^2(at)\Leftrightarrow f(t)=0 ∨...

Thanks, I understood. Another equation is: $$f(t)=\int_0^{\frac{\pi}{a}}f(t)cos(at)dt+1=A+1$$ But we must find f(t), A, B while we have 2 equations?

Homework Statement Find f(x) if:(a∈R) $$f(x)=\int_0^{\frac{\pi}{a}}f(t)cos(at-2ax)dt+1$$ Homework Equations $$f(x)=\int_{a}^{b}f(t)dt=F(a)-F(b)$$ $$\int{udv}=uv-\int{vdu}$$ The Attempt at a Solution I tried to use the fundamental of calculus and integration by parts but they don't have answer...

Thank you.