Circular motion in a magnetic field

AI Thread Summary
The discussion focuses on determining the minimum velocity needed for a negatively charged ball to complete one revolution while suspended in a magnetic field. Key equations include the centripetal force and magnetic force equations, which are combined to analyze the forces acting on the ball at the top of its circular path. The conservation of energy principle is applied to relate the initial and final kinetic and potential energies. The final expressions for velocity are derived, with a specific emphasis on ensuring proper substitutions and squaring terms correctly. The conversation highlights the importance of careful mathematical manipulation in solving physics problems involving circular motion in magnetic fields.
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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?
 
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