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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=2.76×103N,Fz=1.47×103N

Step by step solution

01

Magnetic force

The magnetic force is given by

FB=q(v×B)

02

Determine the magnetic force components acting on that particle

The magnetic force acting on a charged particle is

FB=q(v×B)

Therefore,

F=qv×B=qi^j^k^vxvyvzBxByBz

Therefore, the components of the force are

Fx=q(vyBzvzBy)=0AndFy=q(vzBxvxBz)=qvzBx=(7.26×108C)(5.85×104m/s)(0.650T)=2.76×103N

And,

Fz=q(vxByvyBx)=qvyBx=(7.26×108C)(3.11×104m/s)(0.650T)=1.47×103N

Therefore, the force components areFx=0,Fy=2.76×103N,Fz=1.47×103N

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