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A particle with charge 7.26 * 10-8 C is moving in a region where there is a uniform 0.650-T magnetic field in the +x-direction. At a particular instant, the velocity of the particle has components vx = -1.68 * 104 m/s, vy = -3.11 * 104 m/s, and vz = 5.85 * 104 m/s. What are the components of the force on the particle at this time?

Short Answer

Expert verified

The force components areFx=0,Fy=2.76ร—10โˆ’3N,Fz=1.47ร—10โˆ’3N

Step by step solution

01

Magnetic force

The magnetic force is given by

FBโ†’=q(vโ†’ร—Bโ†’)

02

Determine the magnetic force components acting on that particle

The magnetic force acting on a charged particle is

FBโ†’=q(vโ†’ร—Bโ†’)

Therefore,

Fโ†’=qvโ†’ร—Bโ†’=qi^j^k^vxvyvzBxByBz

Therefore, the components of the force are

Fx=q(vyBzโˆ’vzBy)=0AndFy=q(vzBxโˆ’vxBz)=qvzBx=(7.26ร—10โˆ’8C)(5.85ร—104m/s)(0.650T)=2.76ร—10โˆ’3N

And,

Fz=q(vxByโˆ’vyBx)=โˆ’qvyBx=โˆ’(7.26ร—10โˆ’8C)(โˆ’3.11ร—104m/s)(0.650T)=1.47ร—10โˆ’3N

Therefore, the force components areFx=0,Fy=2.76ร—10โˆ’3N,Fz=1.47ร—10โˆ’3N

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