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An electron experiences a magnetic force of magnitude4.60×10-15Nwhen moving at an angle of 60.0° with respect toa magnetic field of magnitude3.50×10-3T. Find the speed ofthe electron.

Short Answer

Expert verified

The velocity isv=9.49×106m/s

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 speed of the electron

We know that the magnetic field force is given by

F=q(v×B)F=qvBsinθ

Rearranging we get

VFqBsinθ

Plug the values

V=4.60×10-15N1.6×10-19C3.50×10-3Tsin60v=9.496m/sHence,thevelocityisv=9.49×106m/s

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