What is the speed of the electrons when the simultaneous action of an electric f

Question

What is the speed of the electrons when the simultaneous action of an electric field of magnitude 4 times 10^4 v/m and a magnetic field of magnitude 6 times 10^-3T (both fields are perpendicular to the direction of the electron beam and to each other) produces no deflection of the electrons? A straight vertical wire carries a current 10A downwards in the region where a uniform magnetic field of magnitude 0.67 is produced by an electromagnet. What are the magnitude and direction of the magnetic force on a 1cm long section of the wire if the magnetic field direction is a) east; b) south; c) 30degree south of east?

Explanation / Answer

Fe = q*E = 1.6*10^-19C * 1.64*10^4 V/m

the force on the electron by the magnetic field is
Fm = qv*B

Fe=Fm
qE = qvB
E = v*B
v = E/B = 4*10^4/(6*10^-3)
v = 6.6*10^6 m/s

b) Current = I = 10 A ( downward )

Magnetic field = B = 0.6 T ( horizontal)

Wire length = L = 1.00 cm =0.01 m

Angle between wire length and magnetic field = = 90 degree

Magnitude of the magnetic force = F = BIL sin

F =0.6*10*0.01*sin 90

Magnitude of the magnetic force=F =0.06 N

The direction of force is horizontal,31 degree west of north.  

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