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...
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)$$
Both equations are:
$$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 ∨...
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...