A particle with charge ? 5.30 nC is moving in a uniform magnetic field B? =?( 1.

Question

A particle with charge ? 5.30 nC is moving in a uniform magnetic field B? =?( 1.25 T )k^. The magnetic force on the particle is measured to be F? =?( 4.00×10?7 N )i^+( 7.60×10?7 N )j^.

Part B Item 3 Calculate the x-component of the velocity of the particle A particle with charge-5.30 nC is moving ina uniform magnetic field B1.25 T )k. The magnetic force on the particle is measured to be F =-(400-10-7 N )i + ( 7.60x10-7 N )j m/s Submit My Answers Give Up Part C Calculate the y-component of the velocity of the particle m/s Submit My Answers Give Up Part D Calculate the scalar product u·F m/s N Submit My Answers Give Up Part E What is the angle between v and F? Give your answer in degrees

Explanation / Answer

F = q(v x B) = qBz[vx(i*k) + vy(j*k)+ vz(k*k)] = qBz[vx(-j) + vy(i)]

part a )

vx = Fy/-qBz = (7.6 x 10^-7)/-(-5.3 x 10^-9 * -1.25) = -114.72 m/s

part b )

vy = Fx/qBz = (-4 x 10^-7) / (-5.3 x 10^-9 * -1.25) = -60.377 m/s

part c )

v.F = vxFx + vyFy + vzFz

vz does not contribute to the force = vz = 0

v.F = Fy*Fx/-qBz + Fx*Fy/qbz = 0

part d )

v.F = vFcostheta

costheta = 0

theta = 90 degree

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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