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When a particle of charge q > 0 moves with a velocity ofat 45.0° from the +x-axis in the xy-plane, a uniform magnetic field exerts a force along the -z-axis (Fig. P27.51). When the same particle moves with a velocitywith the same magnitude asbut along the +z-axis, a forceof magnitude F2 is exerted on it along the +x-axis. (a) What are the magnitude (in terms of q, v1, and F2) and direction of the magnetic field? (b) What is the magnitude ofin terms of F2?

Short Answer

Expert verified
  1. The magnitude and direction of magnetic field isBy=F2v1qy^
  2. The magnitude of isF1isF1=F22

Step by step solution

01

Magnetic force and electric force

The magnetic force is given by

FB=q(v×B)

The electric force is given by

FE=qE

02

Determine the magnetic field

(a)

The magnetic force is given by

FB=q(v×B)

Now the magnetic force is

F2=F2x^v2=v1k^F2x^=qv1k^×B=qByv1x^+qBxv1y^+0z^Bx=0F2=qv1By

Therefore, the magnetic field isBy=F2V1qy^

03

Determine the magnitude of F1→

(b)

The magnitude of F1is

role="math" localid="1668264415915" F1=F1z^v1=v1cos(45)x^+v1sin(45)y^F1z^=qv1×B=qv1sin(45)Bzx^qv1cos(45)Bzy^+(qv1cos(45)Byqv1sin(45)Bx)z^Now,Bz=0Bx=0F1=Byqv1cos(45)F1=Byqv1cos(45)=(F2v1q)qv1cos(45)F1=F2cos(45)=F22

Therefore, the magnitude of isF1isF1=F22

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