A uniform beam of positively charged particle | vedclass

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(A) Let the positively charged beam be at $x = 0$ and the negatively charged beam be at $x = d$. Since the positive charges move in one direction and

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Vedclass · Answer

(A) Let the positively charged beam be at $x = 0$ and the negatively charged beam be at $x = d$. Since the positive charges move in one direction and the negative charges move in the opposite direction,the currents $I_1$ and $I_2$ produced by these beams are in the same direction. Let the distance from the first beam be $x$. The magnetic field due to the first beam is $B_1 = \frac{\mu_0 I}{2\pi x}$ (directed into the page). The magnetic field due to the second beam at distance $(d - x)$ is $B_2 = \frac{\mu_0 I}{2\pi (d - x)}$ (directed out of the page). The net magnetic field is $B = B_1 - B_2 = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x} - \frac{1}{d - x} ight)$. At the midpoint $x = d/2$,$B = \frac{\mu_0 I}{2\pi} \left( \frac{1}{d/2} - \frac{1}{d/2} ight) = 0$. As $x 	o 0$,$B 	o \infty$,and as $x 	o d$,$B 	o -\infty$. This behavior is correctly represented by the graph in option $A$.

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