Question Part Points Submissions Used Question Part Points Submissions Used Ques

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Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used
Question Part Points Submissions Used Question Part Points Submissions Used An alpha particle (charge +2e) and an electron move in opposite directions from the same point, each with the speed of 4.20 times 105 m/s (see the following figure). Find the magnitude and direction of the total magnetic field these charges produce at point P, which is 3.00 nm from each of them.

Explanation / Answer

If you use the Right-Hand-Rule (RHR): both charge actlike conventional charge moving to the right. Point thumb to right, and then fingers of right hand coil around indirection of B to show that B is OUT OF PAGE.
Starting with Biot-Savart Law: one can replace I with Q / t and then the ds in the numerator of the law multiplied by Q/t gives Q v, where v is now the speed of the moving charge Q. thus, B = (?o/4?) Q v sin? / d2 since ?o = 4? 10^-7, this simplifies to: B = 10^-7 * Q v sin? / d2
B (electron) = 10^-7 * (1.60*10^-19) (4.2*10^5) sin140 / (3*10^-9)2 = 4.8 * 10^-4 T
B(alpha) is same as above eqn for electron, except that Q is doubled. Thus, Ba = 9.6 * 10^-4 T Adding these two results to get the total B field at point P gives B = 1.44 * 10^-3 T
B (electron) = 10^-7 * (1.60*10^-19) (4.2*10^5) sin140 / (3*10^-9)2 = 4.8 * 10^-4 T
B(alpha) is same as above eqn for electron, except that Q is doubled. Thus, Ba = 9.6 * 10^-4 T Adding these two results to get the total B field at point P gives B = 1.44 * 10^-3 T

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