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A particle with charge-5.60nCis moving in a uniform magnetic fieldrole="math" localid="1655717557369" B=-(1.25T)k^

The magnetic force on the particle is measured to berole="math" localid="1655717706597" F=-(3.40×10-7N)i^-(7.40×10-7N)j^ (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there
components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar productv֏F. What is the angle between velocity and force?

Short Answer

Expert verified

(a) The velocity along x axis is Vx=-106m/sand along y axis is Vy=-48.6mls

(b) The z component is not determined, since there is no force along this component hence measurement doesn’t tell us anything about Vz

(c) The angle between the two is 90 degrees

Step by step solution

01

The significance of magnetic field

The magnetic field force is given by

FB=q(v×B)F=qvBsinθ

Where q is the charge of the particle, V is the velocity and B is the magnetic field

02

Determination of the components of the velocity of the particle 

We know that the magnetic field force is given by

F=q(v×B)

We know

F=q֏detijkvxvyvzBxByBz

role="math" localid="1655718671037" Fxi^+Fyj^+Fzk^=qvyBz-vzByi^+qvzBx-vxBzj^+qvxBy-vyBxk^

We know thatBy=Bx=0

Applying this in the equation we get

Fxi^+Fyj^+Fzk^=qvyBzi^-qvxBzj^

Equating the two sides we get

Fx=qvyBzFy=-qvxBzFz=0

Plugging in the values we get

vx=Fy-qBz=7.40×10-7N-5.60×10-9C-1.25T=-106m/sVy=FxqBz=-3.40×10-7N5.60×10-9C-1.25T=-48.6m/s

Hence, the velocity along x axis isvx=-106m/s and along y axis isvy=-48.6m/s

03

Determination scalar Product

We see

v֏F=vxFx+VyFy+VzFzv֏F=Fy-qBzFx+FxqBzFyv֏F=0

This means that the two vectors are perpendicular to each other, so the angle is 90 degrees

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