Why i have forces vectors up and down

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http://i32.tinypic.com/rclu8z.gif

they are supposed to be inide and outside the page

come come they go to be up and down?

(you can left click in order to zoom)
 
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electron2 said:
http://i32.tinypic.com/rclu8z.gif

they are supposed to be inide and outside the page

come come they go to be up and down


?

(you can left click in order to zoom)

The forces on the moving charges in the wires will be orthogonal to both the B-field direction and the direction of the motion of the charges. Hence, the forces have to be up and down in that diagram, since both B and the current directions lie in the plane of the page.

Use the equation F = q v X B to help you get a feel for the directions involved. Use the right-hand rule for that cross product in order to get the correct force direction. Does that make sense now?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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