The Magnitude and Direction of the Force on a Charge Moving in a Magnetic Field
- When a charged particle moves through a magnetic field , it experiences a force known as the magnetic force.
- This force depends on the charge, velocity, magnetic field strength, and the angle between the velocity and the magnetic field.
The Magnitude of the Magnetic Force
The magnitude of the magnetic force on a moving charge is given by the equation:
$$F = qvB \sin \theta$$
where:
- $F$ is the magnetic force (in newtons, N)
- $q$ is the charge of the particle (in coulombs, C)
- $v$ is the speed of the particle (in meters per second, m/s)
- $B$ is the magnetic field strength (in teslas, T)
- $\theta$ is the angle between the velocity vector and the magnetic field vector
- The force is maximized when the charge moves perpendicular to the magnetic field ($\theta = 90^\circ$), as $\sin 90^\circ = 1$.
- If the charge moves parallel to the field ($\theta = 0^\circ$ or $180^\circ$), the force is zero because $\sin 0^\circ = 0$.
The Direction of the Magnetic Force
The direction of the magnetic force is determined by the right-hand rule:
- Point your fingers in the direction of the velocity ($\vec{v}$).
- Curl your fingers toward the direction of the magnetic field ($\vec{B}$).
- Your thumb points in the direction of the force ($\vec{F}$) for a positive charge .
- For a negative charge , the force is in the opposite direction.
Otherwise, you can also use Fleming's left-hand rule:
- Stretch out your thumb, forefinger, and middle finger so that they are mutually perpendicular.
- Point your forefinger in the direction of the magnetic field ($\vec{B}$).
- Point your middle finger in the direction of the current ($\vec{I}$).
- Your thumb then points in the direction of the force ($\vec{F}$) acting on the conductor.
- A proton ($q = +1.6 \times 10^{-19} \text{ C}$) moves at $2 \times 10^6 \text{ m s}^{-1}$ perpendicular to a magnetic field of $0.5 \text{ T}$.
- The force on the proton is:
$$F = qvB \sin \theta = (1.6 \times 10^{-19} \text{ C})(2 \times 10^6 \text{ m/s})(0.5 \text{ T}) \sin 90^\circ = 1.6 \times 10^{-13} \text{ N}$$
The Magnitude and Direction of the Force on a Current-Carrying Conductor in a Magnetic Field
- A current-carrying conductor in a magnetic field also experiences a force.
- This force is the result of the magnetic forces acting on the moving charges (electrons) within the conductor.
The Magnitude of the Magnetic Force
The magnitude of the force on a straight conductor of length $L$ carrying a current $I$ in a magnetic field $B$ is given by:
$$F = BIL \sin \theta$$
where:
- $F$ is the magnetic force (in newtons, N)
- $B$ is the magnetic field strength (in teslas, T)
- $I$ is the current (in amperes, A)
- $L$ is the length of the conductor in the field (in meters, m)
