6. Force on a moving charge in uniform magnetic field
Force on a moving charge in a uniform magnetic field
If a charge \( q \) moves with a velocity \(v\) in a zone of uniform magnetic field \(\vec{B}\), it experiences a force given by:
\[ \vec{F_B} = q \vec{v} \times \vec{B} \]
\[ \vec {F_B} = q \ v\ B \sin \theta \ \hat n\]
This force is called as magnetic force.
If velocity and magnetic field are parallel or anti-parallel, there is no magnetic force on the charge.
Definition of SI unit of magnetic field
SI unit of magnetic field is Tesla.
A magnetic field is 1 Tesla if a charge of 1 Coulomb moving with a speed of 1 m/s in a direction perpendicular to the existing magnetic field experiences a force of 1 N.
Force on a moving charge in uniform magnetic and electric fields
Lorentz force
The force exerted on a charged particle moving through a zone of electric field as well as magnetic field is called as Lorentz force. The force is the vector sum of electric force and magnetic force.
If \(\vec{v}\) is the velocity of charge \( q \) and \(\vec{E}\) and \(\vec{B}\) are electric and magnetic fields respectively, then:
\[ \vec{F} = q (\vec{E} + \vec{v} \times \vec{B}) \]
Where, \(\vec{F}_E = q \vec{E}\) is Electric force in the direction of electric field.
\(\vec{F}_B = q (\vec{v} \times \vec{B})\) is Magnetic force in the direction of the cross product of \(\vec{v}\) and \(\vec{B}\) (using the right-hand rule).
Motion of a charge in a perpendicular uniform magnetic field
Consider a charged particle having charge \( q \) and mass \( m \) moving in a uniform magnetic field. The particle has velocity from left to right, and the magnetic field \( \vec{B} \) is directed away from the observer. The charge \( q \) experiences a force \( \vec{F} = q (\vec{v} \times \vec{B}) \) in the direction upwards. The force on the charge is normal to its velocity, so no work is done by the magnetic force (i.e., kinetic energy remains unchanged). However, there is a change in momentum as the direction of velocity changes.
The magnetic field transfers only momentum, not energy. The electric field transfers both momentum and energy.
Equations and properties
\( F_B = q (\vec{v} \times \vec{B}) \). If \( \vec{v} \) and \( \vec{B} \) are perpendicular:
\[ F_B = qvB \]
\( F_B \) acts as a centripetal force that rotates the charged particle in a circular motion:
\[ qvB = \frac{mv^2}{r} \]
\[r = \frac{mv}{qB}\]
where \( r \) is the radius of the circular path and \(m\) is the mass of the charged particle.
If \( \omega \) is the angular velocity and \( \nu \) is the frequency of rotation:
\[ v = \omega r \] and
\[ \omega = \frac{qB}{m} \]
\[\nu = \frac{qB}{2 \pi m}\]
Frequency of rotation is independent of energy and velocity of the charge.
Motion of charge when its velocity is not perpendicular to its magnetic field
Consider a charge \( q \) moving with velocity \( \vec{v} \) in a magnetic field region. The angle between \( \vec{v} \) and \( \vec{B} \) is \( \theta \). Velocity \( \vec{v} \) can be resolved along \(\vec{B}\) and perpendicular to \(\vec{B}\). Due to the perpendicular component of velocity, the charge has a circular motion with radius \( r = \frac{mv_\perp}{qB} \) (as discussed above) and a linear motion along \( \vec{B} \). The two motions produce a helical motion. The radius \( r \) is the radius of the helix. Frequency of the motion is \( \nu = \frac{qB}{2\pi m} \). Time period \( T = \frac{2\pi m}{qB} \). Displacement along \( \vec{B} \) by charge in one cycle \(pitch\) \( P = v_\parallel T = \frac{2\pi m v_\parallel}{qB} \).