zaz A proton of mass m is projected into a uniform field B = Ba₂ with an initial velocity aa + ßa₂. (a) Find the differential equations that the position vector r = xax + yay + must satisfy. (b) Show that a solution to these equations is -sin oot, y - cos wt, z = ßt where w = eB/m and e is the charge on the proton. (c) Show that this solution describes a circular helix in space. x dx Answer: (a)- = a cos wt, dt dy dt = -a sin wt, dz - = dt ß, (b) and (c) Proof.

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A proton of mass \( m \) is projected into a uniform field \( \mathbf{B} = B_0 \mathbf{a_z} \) with an initial velocity \( \alpha \mathbf{a_x} + \beta \mathbf{a_z} \). 

**(a)** Find the differential equations that the position vector \( \mathbf{r} = x \mathbf{a_x} + y \mathbf{a_y} + z \mathbf{a_z} \) must satisfy.

**(b)** Show that a solution to these equations is
\[
x = \frac{\alpha}{\omega} \sin \omega t, \quad y = \frac{\alpha}{\omega} \cos \omega t, \quad z = \beta t
\]
where \( \omega = \frac{eB_0}{m} \) and \( e \) is the charge on the proton.

**(c)** Show that this solution describes a circular helix in space.

**Answer:**

**(a)** The differential equations are:
\[
\frac{dx}{dt} = \alpha \cos \omega t, \quad \frac{dy}{dt} = -\alpha \sin \omega t, \quad \frac{dz}{dt} = \beta
\]

**(b)** and **(c)** Proof.
Transcribed Image Text:A proton of mass \( m \) is projected into a uniform field \( \mathbf{B} = B_0 \mathbf{a_z} \) with an initial velocity \( \alpha \mathbf{a_x} + \beta \mathbf{a_z} \). **(a)** Find the differential equations that the position vector \( \mathbf{r} = x \mathbf{a_x} + y \mathbf{a_y} + z \mathbf{a_z} \) must satisfy. **(b)** Show that a solution to these equations is \[ x = \frac{\alpha}{\omega} \sin \omega t, \quad y = \frac{\alpha}{\omega} \cos \omega t, \quad z = \beta t \] where \( \omega = \frac{eB_0}{m} \) and \( e \) is the charge on the proton. **(c)** Show that this solution describes a circular helix in space. **Answer:** **(a)** The differential equations are: \[ \frac{dx}{dt} = \alpha \cos \omega t, \quad \frac{dy}{dt} = -\alpha \sin \omega t, \quad \frac{dz}{dt} = \beta \] **(b)** and **(c)** Proof.
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