A potential difference of magnitude V accelerates a parade of mass m and charge

Question

A potential difference of magnitude V accelerates a parade of mass m and charge +q from rest to the right, shown. The potential on the right is (higher, lower) than the potential on the left (Please circle one ) Determine the magnitude of the velocity of the parade after it passes through the potential difference, in terms of any or all of the following V, q, m After passing through the potential difference, the parade enters a uniform magnetic field, and moves in a circular path of radius r, as shown The direction of this magnetic field must be (right, left, up down, in, out) (Please circle one). Beginning with Newton's Second law ( = ma) show that the radius of the circular path of the charged particle is given by mv/qB, where v is the magnitude of the velocity of the particle.

Explanation / Answer

(a) As we know that , the positve charge will moves from High potenatil difference to low potenatial diffrence thus, potential on the right is LOWER , whereas potenatil on the left is HIGH (b) Acc. to conservation of energy we have , POtenatial difeerence applied to the particle is equal to kinetic energy is attained in the particle thus, potenatil energy : q V Kinetic energy : 1/2 m v2 thus, q V = 1/2mv2         Velocity of the particle ;               v = [ 2qV /m ] (c) Since , the particle is tranversed in clock wise direction , the direction of the magnetic field should be INWARD (d) Mangetic force on the particle is given by the formual as Fb = q v B   When it moves in circular path , centripetal force acting on it Fc = m a      where a is the centri=petla accelration : a = v2 / r thus, from Nweton's second law Fc = ma   = mv2 / r mv2/ r = q vB thus, radius of the particle    r = mv / B q (d) Mangetic force on the particle is given by the formual as Fb = q v B   When it moves in circular path , centripetal force acting on it Fc = m a      where a is the centri=petla accelration : a = v2 / r thus, from Nweton's second law Fc = ma   = mv2 / r mv2/ r = q vB thus, radius of the particle    r = mv / B q

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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