So I have A done already it equals 3.37x10^-6 N and it\'s 30 degrees below x-axi

Question

So I have A done already it equals 3.37x10^-6 N and it's 30 degrees below x-axis. I NEED HELP ON PART B BOTH (i) AND (ii) thank you!! Three points (A, B, and C) make up the vertices ofa right triangle. The distance between points A and B is 4.0 cm. Two currents are oriented have equal to the paper and are placed at points A and B. The currents magnitudes of 0.45 A, but point in opposite directions. A. What is the net magnetic field at point C? Give both its magnitude and its direction with respect to the axes shown. 30 Menu B. Suppose that a particle with a mass of2.07 x 10 kg and a charge of-1.5 pc is passing through point C with a velocity of magnitude 3.4 x 10 m/s pointed at an angle off above the +x-axis. 15 of the force. (i) the force.

Explanation / Answer

magnetic field from part A i s B = 3.37*10^-6 N,30 degrees below x axis

now the particle of mass m = 2.07*10^-8 kg, velocity v = 3.4*10^3 m/s, at angle 15 degrees above the x axis
charge of particle q = -1.5*10^-6 C

so the net angle theta = 30+15 = 45 degrees , is the angle between the magnetic field and velocity

now the magnetic force is
i)
   F = qvB sin theta

   = (-1.5*10^-6)(3.4*10^3)(3.37*10^-6)sin45 N = (-1.21530)*10^-8 N

the direction is into the page

ii) instantaneous acceleration is F = ma ==> a = F/m

  

   a = (-1.21530)*10^-8/(2.07*10^-8) = - 0.5871 m/s2

acceleration is also in the direction of the force that is into the page

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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