Circular motion in a magnetic field

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kaspis245
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Homework Statement


Small mass ##m## ball has a negative charge ##q## and is hanging on an inelastic string which has a length of ##l##. What is the smallest velocity that one need to impart on this ball for it to make one revolution? There is also a uniform magnetic field ##B## as shown in the drawing.
Circular_motion.png


Homework Equations


##F=Bqv##
##F_c=\frac{mv^2}{r}##

The Attempt at a Solution


We need to find ##v_o##

Conservation of energy:
(1) ##\frac{mv_o^2}{2}=2mgl+\frac{mv^2}{2}##

The force ##F=Bqv## always points into the center of the circle. When the ball reaches the top of the circle, it will be affected by two forces ##F=Bqv## and ##mg##. Both point downwards, hence the sum of those forces must be the centripetal force.
(2) ##Bqv+mg=\frac{mv^2}{l}##

Now I can express v from this equation and place it into (1). Is that correct?
 
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kaspis245 said:

Homework Statement


Small mass ##m## ball has a negative charge ##q## and is hanging on an inelastic string which has a length of ##l##. What is the smallest velocity that one need to impart on this ball for it to make one revolution? There is also a uniform magnetic field ##B## as shown in the drawing.
Circular_motion.png


Homework Equations


##F=Bqv##
##F_c=\frac{mv^2}{r}##

The Attempt at a Solution


We need to find ##v_o##

Conservation of energy:
(1) ##\frac{mv_o^2}{2}=2mgl+\frac{mv^2}{2}##

The force ##F=Bqv## always points into the center of the circle. When the ball reaches the top of the circle, it will be affected by two forces ##F=Bqv## and ##mg##. Both point downwards, hence the sum of those forces must be the centripetal force.
(2) ##Bqv+mg=\frac{mv^2}{l}##

Now I can express v from this equation and place it into (1). Is that correct?
Yes. That looks to be correct.
 
Please check if my final answer is correct.

##mv^2-Bqlv-mgl=0##
##v=\frac{Bql+\sqrt{B^2q^2l^2+4m^2gl}}{2m}##

##v_o=\sqrt{4lg+\frac{Bql+\sqrt{B^2q^2l^2+4m^2gl}}{2m}}##
 
kaspis245 said:
Please check if my final answer is correct.

##mv^2-Bqlv-mgl=0##
##v=\frac{Bql+\sqrt{B^2q^2l^2+4m^2gl}}{2m}##
I think this expression for ##v## is correct.

##v_o=\sqrt{4lg+\frac{Bql+\sqrt{B^2q^2l^2+4m^2gl}}{2m}}##

Did you forget to square ##v## when you substituted for ##v## inside the radical?