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A particle with charge 7.80 μC is moving with velocity v=-(3.80×103m/s)j^. The magnetic force on the particle is measured to be (7.60×10-3N)i^-(5.20×10-3N)k^.(a) Calculate all the components of the magnetic field you canfrom this information. (b) Are there components of the magneticfield that are not determined by the measurement of the force?Explain. (c) Calculate the scalar productBF what is the angle between Bands F?

Short Answer

Expert verified
  1. Magnetic field along x axis isBx=-0.175T and along z axis isBz=-0.256T
  2. The y component is not determined, since there is no force along this component hence measurement doesn’t tell us anything aboutBy

c. The angle between the two is 90 degrees

Step by step solution

01

The significance of the 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 magnetic field

We know that the magnetic field force is given by

F=q(v×B)

We know

F=qdeti^j^k^vxvyvzBxByBzFxi^+Fyj^+Fzk^=qvyBzvzByi^+qvzBxvxBzj^+qvxByvyBxk^

We knowVx=0 thatVz=0

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

Comparing we get

Fx=qvyBzFy=0Fz=qvyBx

Substitute all the value in the above equation.

Solve for the magnetic field components

Bz=Fxqvy=7.60×103N7.80×106C×3.80×103m/s=0.256TBx=Fzqvy=5.20×103N7.80×106C×3.80×103m/s=0.175T

Hence, Magnetic field along x axis isBx=-0.175T and along z axis is Bz=-0.256T.

03

Determination of the scalar product

We see

BF=BxFx+ByFy+BzFzBF=(0.175T)7.60×103N+(0.256T)5.20×103NBF=0

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

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Most popular questions from this chapter

When switch Sin Fig. E25.29 is open, the voltmeter V reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery, and the circuit resistance R. Assume that the two meters are ideal, so they don’t affect the circuit.

Fig. E25.29.

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