91Ó°ÊÓ

Kinetic energy is the energy that an object possesses due to its motion. In the context of an \(\alpha\)-particle, kinetic energy is gained from being accelerated through a potential difference, essentially converting electric potential energy into kinetic energy. Think of potential difference like a slide; the steeper the slide (higher the potential difference), the faster the particle moves at the bottom, gaining kinetic energy.The equation to calculate the kinetic energy \(K\) from a potential difference \(V\) is \(K = qV\), where \(q\) represents the charge of the particle. Here, the charge \(q\) equals \(+2e\), because an \(\alpha\)-particle carries two positive charges (which follows from its composition of two protons). By substituting the charge of an electron \(e = 1.60 \times 10^{-19} \, \text{C}\) and the potential difference \(1.20 \times 10^{6} \, \text{V}\), we calculate the kinetic energy as \(3.84 \times 10^{-13} \, \text{J}\).Knowing the kinetic energy helps us find the speed of the particle by relating it to the particle's mass \(m\) using \(\frac{1}{2} mv^2 = K\). This demonstrates how potential energy is transformed into kinetic energy, setting the particle into motion.

When an \(\alpha\)-particle enters a magnetic field, it experiences a force that makes it move in a circular path. This is due to the magnetic force acting perpendicular to the motion of the particle, continually changing the direction of the particle's velocity while maintaining a constant speed.The constant circular path means the particle is in uniform circular motion, which occurs when an object moves in a circle at a constant speed. In this scenario, the centripetal force necessary to keep the particle in circular motion is provided by the magnetic force. Therefore, knowing the speed \(v\), mass \(m\), and magnetic field \(B\), we can find the radius \(r\) of the circular path using the formula \(r = \frac{mv}{qB}\).In the exercise, substituting the known values \(m = 6.64 \times 10^{-27} \, \text{kg}\), \(v = 1.08 \times 10^{7} \, \text{m/s}\), \(q = 2 \times 1.60 \times 10^{-19} \, \text{C}\), and \(B = 2.20 \, \text{T}\), we find \(r \approx 2.05 \times 10^{-2} \, \text{m}\). This radius gives the size of the path our particle traverses, forming a neat circle in the magnetic field's presence. This showcases the neat interplay between velocity, magnetic force, and circular motion.

Magnetic force is the key player acting on charged particles when they encounter a magnetic field. This force causes the charged particle to move in a circular trajectory, as seen with the \(\alpha\)-particle in our exercise.The magnetic force \(F\) experienced by a charged particle moving through a magnetic field can be calculated using the formula \(F = qvB \sin \theta\). When the particle's motion is perpendicular to the field, as is the case in our scenario, \(\sin \theta = 1\) because \(\theta = 90^\circ\). Thus, \(F = qvB\).The exercise results in \(F = 7.61 \times 10^{-12} \, \text{N}\) when applying the known values. This force constantly redirects the \(\alpha\)-particle, creating a balanced path that traces a circle. This relationship elegantly links the particle's velocity, the magnetic field, and the force experienced, illustrating the control that magnetic fields exert over charged particles.