Circular motion in a magnetic field

In summary: Yes, I forgot to square it.Thank you for catching that.In summary, if you impart a velocity of ##v_o## on the small mass ball, it will make one revolution.
  • #1
kaspis245
189
1

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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  • #2
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.
 
  • #3
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}}##
 
  • #4
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?
 

Related to Circular motion in a magnetic field

What is circular motion in a magnetic field?

Circular motion in a magnetic field is the movement of a charged particle in a circular path due to the interaction between the magnetic field and the particle's charge.

What is the formula for calculating the radius of a particle's circular motion in a magnetic field?

The formula for calculating the radius of a particle's circular motion in a magnetic field is r = mv/qB, where r is the radius, m is the mass of the particle, v is the velocity, q is the charge of the particle, and B is the strength of the magnetic field.

What is the direction of the force on a particle in circular motion in a magnetic field?

The direction of the force on a particle in circular motion in a magnetic field is perpendicular to both the magnetic field and the velocity of the particle. It follows the right-hand rule, where the thumb points in the direction of the velocity, the fingers point in the direction of the magnetic field, and the force is directed outwards from the palm.

How does the strength of the magnetic field affect the radius of the particle's circular motion?

The strength of the magnetic field is directly proportional to the radius of the particle's circular motion. This means that as the magnetic field increases, the radius of the circular path also increases, and vice versa.

What is the relationship between the charge of the particle and the radius of its circular motion in a magnetic field?

The charge of the particle is inversely proportional to the radius of its circular motion. This means that as the charge increases, the radius decreases, and vice versa. This relationship can also be seen in the formula for calculating the radius, where q is in the denominator.

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