A. What is the net magnetic field at point C? Give both it\'s magnitude and its

Question

A. What is the net magnetic field at point C? Give both it's magnitude and its direction with respect to the axes shown. B. Suppose that a particle with a mass of 2.07 E -8 kg and a charge of -1.5 uC is passing through point C with a velocity of magnitude 3.4 E 3 m/s pointed at an angle of 15 degrees above the +x-axis I. What is the net magnetic force felt by the particle? Give both direction and magnitude of the force? II. What is the instantaneous acceleration of the particle? Again give both the magnitude and do of the forc. distance between points A and B is 4.0 cm. Two currents are oriented perpendicular to the paper and are placed at points A and B. The currents have equal magnitudes of 0.45 A, but point in opposite directions. A. What is the net magnetic field at point C? Give both its magnitude 30° and its direction with respect to the axes shown. I 15A a ME ppose that a particle with a mass of 2.07 10 kg and a charge HC through int C with a velocity of magnitude 3.4 x 10' at an angle of 15 above the +x-axi m/s pointed What is the net magnetic force felt by the particle? Give both the magnitude and the direction ofthe force. 2016 (i) What is the instantaneous acceleration of the particle? Again, give both the magnitude and the direction of the force.

Explanation / Answer

1) magnetic field due to A,

BAx = uo*I*cos30/(2pi*a)) = 4*pi*10^-7*0.45*cos30/(2*pi*0.04) = 1.95*10^-6 T

BAy = -uo*I*sin30/(2pi*a)) = -4*pi*10^-7*0.45*sin30/(2*pi*0.04) = -1.125*10^-6 T

magnetic field due to B,

BBx = uo*I*cos30/(2pi*a)) = 4*pi*10^-7*0.45*cos30/(2*pi*0.04) = 1.95*10^-6 T

BBy = +uo*I*sin30/(2pi*a)) = 4*pi*10^-7*0.45*sin30/(2*pi*0.04) = 1.125*10^-6 T

Bx = BAx + BBx = 3.9*10^-6 T

By = BAy + BBy = 0

B = sqrt(Bx^2+By^2) = 3.9*10^-6 T (Ans)

2) i) magnetic force Fb = q*(v x B)

v = 3.4*10^3*cos15i + 3.4*10^3*sin15j

v = 3.28*10^3 i + 0.877*10^3 j

B = 3.9*10^-6 i

Fb = 1.5*10^-6*((3.28*10^3 i + 0.877*10^3 j) x 3.9*10^-6 i )

Fb = -5.130*10^-9 N k

magnitude = 5.130*10^-9 Nk

direction along -z axis

ii) acceleration a = F/m = 5.130*10^-9/(2.07*10^-8) = 0.24 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.