Question

A charged particle is moving perpendicular to a uniform magnetic field. What will be the path of the

Answer 1

Circle

Answer 2

• Lorentz Force: When a charged particle with charge $q$ moves with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$, it experiences a magnetic force given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$

• Direction of Force: Since the particle is moving perpendicular to the magnetic field ($\vec{v} \perp \vec{B}$), the angle between $\vec{v}$ and $\vec{B}$ is $90^\circ$. The magnitude of the force is $F = |q|vB \sin(90^\circ) = |q|vB$. The direction of this force is always perpendicular to both $\vec{v}$ and $\vec{B}$.

• Work Done by Magnetic Force: Because the magnetic force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$ of the particle, the work done by the magnetic force on the particle is zero ($W = \vec{F} \cdot \vec{v} dt = 0$). This implies that the kinetic energy of the particle remains constant, and thus its speed ($|\vec{v}|$) remains constant.

• Centripetal Force: A force that is always perpendicular to the velocity and has a constant magnitude, causing the particle to move with constant speed, acts as a centripetal force. This force continuously changes the direction of the velocity without changing its magnitude.

• Circular Motion: For uniform circular motion, the centripetal force required is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the particle and $r$ is the radius of the circular path. Equating the magnetic force to the centripetal force: $|q|vB = \frac{mv^2}{r}$ Solving for the radius $r$: $r = \frac{mv}{|q|B}$ Since $m$, $v$ (constant speed), $|q|$, and $B$ (uniform field) are all constant, the radius $r$ of the path is constant.

Therefore, the charged particle will follow a circular path.