l Help Sources of Magnetism Begin Date: 3/21/2018 12:01:00 AM -- Due Date: 3/30/
ID: 2031058 • Letter: L
Question
l Help Sources of Magnetism Begin Date: 3/21/2018 12:01:00 AM -- Due Date: 3/30/2018 11:59:00 PM End Date: 6/1/2018 11:59:00 PM (13%) Problem 3: Consider the two charges in the figure, which are moving in opposite directions and located a distance 1 cm on either side of the origin of the given coordinate system. Charge one s 24.5 C and is moving at a speed 11.5 x 10 m/s, and charge two is-7.5 IC and is moving at a speed of 6.5 x 10° m/s. Notice that the positive r-direction in this figure is directed out of the page 12 91 0 what is the total magnetic field, in the-direction, at a point Q located a distance 21 cm down the x-axis, as shown? Your answer should be in tesla Grade Summary 090 100% Potential sin() cotanasinacos0 atan acotan sinh0 cosh tanh) cotanhO cos Submissions Attempts remaining: 8 (3% per attempt) detailed view O Degrees Radians DELI CLEAR Submit I give up! Hints: 3 for a 0% deduction. Hints! remaining: 0 Feedback: 1% deduction per feedback. How do you find the magnetic field due to a moving point charge? -Use The Biot-S the ficlds. What happens to the direction of the field produced if the charge is negative? Submission History avart law; paying careful attention to the direction of Answer Hints Feedback Totals Totals 0% 0%Explanation / Answer
we know,
magnetic field due to a moving charged particle,
B = (mue/(4*pi))*(q*v*sin(theta))/r^2))
so, B1 = (mue/(4*pi))*(q1*v1*sin(theta))/r^2))
= 10^-7*24.5*10^-6*11.5*10^6*(d/sqrt(d^2 + xo^2))/(d^2 + xo^2)
= 10^-7*24.5*10^-6*11.5*10^6*d/(d^2 + xo^2)^1.5
= 10^-7*24.5*10^-6*11.5*10^6*0.01/(0.01^2 + 0.21^2)^1.5
= 3.03*10^-5 T (towards +z axis)
B2 = (mue/(4*pi))*(q2*v2*sin(theta))/r^2))
= 10^-7*7.5*10^-6*6.5*10^6*(d/sqrt(d^2 + xo^2))/(d^2 + xo^2)
= 10^-7*7.5*10^-6*6.5*10^6*d/(d^2 + xo^2)^1.5
= 10^-7*7.5*10^-6*6.5*10^6*0.01/(0.01^2 + 0.21^2)^1.5
= 5.25*10^-6 T (towards -z axis)
Bz = B1 - B2
= 3.03*10^-5 - 5.25*10^-6
= 2.51*10^-5 T
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