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A charged particle $q$ enters a uniform magnetic field $\vec{B}$ with velocity $\vec{v}$ making an angle $\theta$ with it. Since the Lorentz force is perpendicular to the velocity, the particle will move along a circular path of radius $r$, which my textbook derives as follows:

$$\frac{mv^2}{r}=qvB \sin\theta$$ $$r=\frac{mv}{qB\sin\theta}.$$

But I think the correct formula for $r$ should be derived as follows:

$$\frac{m(v\sin\theta)^2}{r}=qvB \sin\theta$$ $$r=\frac{mv\sin\theta}{qB}.$$

This should be because we only consider the perpendicular component of velocity when we calculate magnetic force and therefore the velocity to which the force is perpendicular is the component of velocity perpendicular to $\vec{B}$ and not $\vec{v}$.

Which is the correct formula?

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Your derivation is correct and your book is incorrect unless the $v$ in their equation is the component of velocity perpendicular to the magnetic field?
The diagram below assumes a positive charge.

enter image description here

The radius of the circular motion is given by the equation $r=\dfrac{mv\sin\theta}{qB}$ and the pitch of the helix is $p = \dfrac{2\pi mv\cos \theta}{qB}$

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    $\begingroup$ Why then does the particle describe helical motion? The velocity at any point in this case would not be parallel to the plane of circular motion. $\endgroup$ – Omar Abdullah Feb 13 '17 at 16:31
  • $\begingroup$ What makes you think that the motion is helical as the only force on the charge is the one that produces the centripetal acceleration of the charge? $\endgroup$ – Farcher Feb 13 '17 at 16:32
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    $\begingroup$ If the particle has a component of its motion along the field direction, that motion is constant, since there can be no component of the magnetic force in the direction of the field. The general motion of a particle in a uniform magnetic field is a constant velocity parallel to $\vec{B}$ and a circular motion at right angles to $\vec{B}$—the trajectory is a cylindrical helix. - Feynman Lectures. Where do I misunderstand this? $\endgroup$ – Omar Abdullah Feb 13 '17 at 16:47
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    $\begingroup$ @OmarAbdullah I am sorry. You (and Feynman) are correct and I have amended my answer. $\endgroup$ – Farcher Feb 13 '17 at 17:18
  • $\begingroup$ The $\vec{v}$ in the equation in my book is the actual velocity. Thanks for confirmation. $\endgroup$ – Omar Abdullah Feb 13 '17 at 18:01
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enter image description here.this is the ans for the question

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    $\begingroup$ Please type out your answer, rather than just posting a picture. $\endgroup$ – Chris Jul 21 at 21:55

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