To understand the behavior of a charged particle moving through a magnetic field, we need to break down the problem into manageable parts. When a charged particle is accelerated through a potential difference, it gains kinetic energy, which can be expressed in terms of its mass and velocity. Once it enters a magnetic field, the force acting on it causes it to move in a circular or helical path, depending on its initial velocity and the orientation of the magnetic field.
Acceleration Through Potential Difference
When a particle with charge \( q \) is accelerated through a potential difference \( V \), it gains kinetic energy given by the equation:
Kinetic Energy (KE) = qV
This kinetic energy can also be expressed in terms of the particle's mass \( m \) and velocity \( v \):
KE = \frac{1}{2} mv^2
Setting these two expressions for kinetic energy equal gives us:
qV = \frac{1}{2} mv^2
From this, we can solve for the velocity \( v \):
v = \sqrt{\frac{2qV}{m}}
Motion in a Magnetic Field
Once the particle enters a uniform magnetic field \( B \) that is perpendicular to its direction of motion, it experiences a magnetic force given by:
F = qvB
This magnetic force acts as a centripetal force, causing the particle to move in a circular path. The centripetal force required for circular motion is:
F = \frac{mv^2}{r}
Setting these two forces equal gives:
qvB = \frac{mv^2}{r}
From this equation, we can solve for the radius \( r \) of the circular path:
r = \frac{mv}{qB}
Substituting \( v \) from our earlier equation, we find:
r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2Vm}{q}} \cdot \frac{1}{B}
Helical Motion of the Particle
Now, let's consider the specific case where a particle with a de Broglie wavelength of \( 2.21 \times 10^{-13} \) m and charge \( 1.6 \times 10^{-19} \) C is projected at a speed of \( 2 \times 10^6 \) m/s at an angle of \( 60^\circ \) to the x-axis in a magnetic field of \( 0.3 \) T along the y-axis.
Calculating the Radius
The radius of the helical path can be calculated using the formula derived earlier. First, we need to find the component of the velocity that is perpendicular to the magnetic field:
v_{\perp} = v \sin(60^\circ) = 2 \times 10^6 \cdot \frac{\sqrt{3}}{2} \approx 1.732 \times 10^6 \text{ m/s}
Now, substituting into the radius formula:
r = \frac{mv_{\perp}}{qB}
Assuming the mass \( m \) can be derived from the de Broglie wavelength using the relation:
\lambda = \frac{h}{mv} \Rightarrow m = \frac{h}{\lambda v}
Using \( h \approx 6.626 \times 10^{-34} \) J·s, we can find \( m \) and then substitute it back to find \( r \). After calculations, we find:
r \approx 0.03 \text{ m}
Determining the Time Period
The time period \( T \) of the helical motion can be calculated using the formula:
T = \frac{2\pi m}{qB}
Substituting the values gives:
T \approx 6.25 \times 10^{-8} \text{ s}
In summary, the charged particle moves in a helical path with a radius of approximately \( 0.03 \) m and a time period of about \( 6.25 \times 10^{-8} \) seconds due to the influence of the magnetic field. This behavior illustrates the fascinating interplay between electric and magnetic forces on charged particles.
