Physics: Consider a circular conducting wire of radius r carrying a constant current i. the loop lies in the zy-plane and the center of the loop is at origin of the coordinates. the z-axis goes through the center of the loop and it is perpendicular to the plane where the loop lies. the current in the loop flows counterclockwise when viewed from a point on the 2-axis, with 2 0 a) from the law of biot-savart, show that the magnetic field on the z axis is (10 points) b _ir 2 (2+) tk two current loops, identical

Complete Question The diagram for this question is shown on the first uploaded image The initial angular momentum is Explanation:From the question we are told that The mass of the rod is The mass of the ball is The speed is The angle the ball strikes the rod is The distance from the center of the rod is The length L is The initial angular momentum of the ball is mathematically represented as Substituting the...

Consider a circular conducting wire of radius r carrying

Ответ:

mhurtado143

14.01.2022

Physics

a) it has been shown

b) $B=\frac{\mu_0I}{2}(\frac{R^2}{(R^2+(z-\frac{L}{2})^2)^{\frac{3}{2}}}+\frac{R^2}{(R^2+(z+\frac{L}{2})^2)^{\frac{3}{2}}}) \hat k$

c) it has been shown

d) $\frac{d^2B}{dz^2}=-\frac{3R^2}{(R^2+(-\frac{L}{2}+z)^2)^{\frac{5}{2}}}-\frac{3R^2}{(R^2+(\frac{L}{2}+z)^2)^{\frac{5}{2}}}+\frac{15R^2(-\frac{L}{2}+z)^2}{(R^2+(-\frac{L}{2}+z)^2)^{\frac{7}{2}}}+\frac{15R^2(\frac{L}{2}+z)^2}{(R^2+(\frac{L}{2}+z)^2)^{\frac{7}{2}}}$

e) L/R = 1

Explanation:

a) $dB = \frac{\mu_0}{4\pi}\frac{Idlsin(\theta)}{R^2+z^2}$ where $\theta = 90$ °

Then, $dB=\frac{\mu_0}{4\pi}\frac{Idl}{R^2+z^2}$

$dB_a=dBcos(\phi)$ where $cos(\phi)=R/\sqrt{R^2+z^2}$

Then, $dB_a=\frac{\mu_0}{4\pi}\frac{Idl}{R^2+z^2}\frac{R}{\sqrt{R^2+z^2}}$

So total magnetic field can be found as follows,

$B=\int dB_a=\frac{\mu_0}{4\pi}\frac{I}{R^2+z^2}\frac{R}{\sqrt{R^2+z^2}}\int\limits^{2\pi R}_0 dl=\frac{\mu_0IR}{4\pi(R^2+z^2)^{\frac{3}{2}}}2\pi R=\frac{\mu_0I}{2}\frac{R^2}{(R^2+z^2)^{\frac{3}{2}}} \hat k$

b) Since the first loop is L/2 distance above center, when we choose arbitrary point at z-axis, the distance will be z - L/2. Also, since the second loop is L/2 below the center, same arbitrary point's distance from the loop will be z + L/2. Moreover, since the position of loops is symmetric, choosing arbitrary point from anywhere will not change the result. So now, let's write the above equation down with corresponding consideration.

Since there are 2 loops, their magnetic fields will be added to each other to find the total magnetic field.

Thus,

$B=\frac{\mu_0I}{2}(\frac{R^2}{(R^2+(z-\frac{L}{2})^2)^{\frac{3}{2}}}+\frac{R^2}{(R^2+(z+\frac{L}{2})^2)^{\frac{3}{2}}}) \hat k$

c) The derivative of the equation in part (a) will be as follows,

$\frac{dB}{dz}=-\frac{3R^2(-\frac{L}{2}+z)}{(R^2+(-\frac{L}{2}+z)^2)^{\frac{5}{2}}}-\frac{3R^2(\frac{L}{2}+z)}{(R^2+(\frac{L}{2}+z)^2)^{\frac{5}{2}}}$

At z=0 , the above equation becomes as follows,

$\frac{dB}{dz}=\frac{3R^2\frac{L}{2}}{(R^2+\frac{L^2}{4})^{\frac{5}{2}}}-\frac{3R^2\frac{L}{2}}{(R^2+\frac{L^2}{4})^{\frac{5}{2}}}=0$

d) The second derivative of the equation in part (a) will be as follows,

$\frac{d^2B}{dz^2}=-\frac{3R^2}{(R^2+(-\frac{L}{2}+z)^2)^{\frac{5}{2}}}-\frac{3R^2}{(R^2+(\frac{L}{2}+z)^2)^{\frac{5}{2}}}+\frac{15R^2(-\frac{L}{2}+z)^2}{(R^2+(-\frac{L}{2}+z)^2)^{\frac{7}{2}}}+\frac{15R^2(\frac{L}{2}+z)^2}{(R^2+(\frac{L}{2}+z)^2)^{\frac{7}{2}}}$

e) At z=0 , the above equation becomes as follows,

$-\frac{3R^2}{(R^2+(-\frac{L}{2})^2)^{\frac{5}{2}}}-\frac{3R^2}{(R^2+(\frac{L}{2})^2)^{\frac{5}{2}}}+\frac{15R^2(-\frac{L}{2})^2}{(R^2+(-\frac{L}{2})^2)^{\frac{7}{2}}}+\frac{15R^2(\frac{L}{2})^2}{(R^2+(\frac{L}{2})^2)^{\frac{7}{2}}}=0$