An electron (charge=-1.6 times 10-19 C) is moving at 3 times 10 5 m/s in the pos

Question

An electron (charge=-1.6 times 10-19 C) is moving at 3 times 10 5 m/s in the positive x direction. a magnetic field of 0.8 T is in the positive z direction. What are the magnitude and direction the magnetic force on the electron Electrons (mass me, charge-e)are accelerated from react though a potential difference V and electrons are doing a uniform circular motion. Find out the radius of the trajectory (express your result in terms of m e V and B) A copper wire is bent in to the shape shown in the figure which includes two straight portions and a semi-circle with radius R and center P A current I flows though the wire Find out the magnitude of the magnetic field due to the current flow at the center P.

Explanation / Answer

One answer per posting so here is number 2 2) First the electron is accelerate convervation of energy gives us eV=1/2 mv^2 so v= sqrt(2eV/m) then in the magnetic field we have circular motion so F=mv^2/r F=qvB so qvB=mv^2/r r=mv/qB=m/qb * sqrt(2eV/m)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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