See Our New JEE Book on Amazon

Force on a Moving Charge in a Uniform Magnetic Field


When charged particles move in a magnetic field they experience a force. This force is referred to as the magnetic Lorentz force and acts perpendicular to the motion of the particle. A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force given by \begin{align} \vec{F}_m=q\;\vec{v}\times\vec{B} \end{align}

The magnetic force lies in a plane containing $\vec{v}$ and $\vec{B}$. The magnitude of the force is maximum when magnetic field and velocity are perpendicular and it is zero when magnetic field is parallel to the velocity.


When this topic is taught, the above mentioned charged particles are taken to be electrons, protons or alpha particles and the students have to visualize their motion when they are projected in the magnetic field. In this demonstration ions of an electrolyte are made to move in the magnetic field of a permanent magnet. The design allows the whole liquid mass to move in circular motions which can be seen.


You need a flat bottomed round plastic/glass cap of around 2 inch diameter or a petri dish, crocodile clips, salt water, battery, ring magnet, and copper wires.

  1. Keep the plastic cap on the ring magnet.
  2. Then remove the enamel of about 7 inch long copper wire and put it just inside the periphery of the cap.
  3. Connect the copper wire to the negative terminal of the battery with the help of a crocodile clip. Pour salt water in the plastic cap so that the copper wire gets immersed in it.
  4. Now connect the positive terminal of the battery to a crocodile clip with the help of a copper wire.
  5. Hold this crocodile clip vertically in the centre of the plastic cap such that a small metal portion of the clip is inside the salt water. Switch on the battery (9 V).

See that the salt water in the plastic cap gradually starts rotating in clockwise direction. After a while the water becomes dark and the rotation is much more visible. Interchange the terminals of the battery. See that the water now starts rotating in anticlockwise direction. Now invert the ring magnet under the cap, water starts rotating again in clockwise direction.


The arrangement made in the plastic cap makes it a voltameter. The salt water acts as the electrolyte. Copper wire which is at the periphery and the metal portion of the crocodile clip, act as the two electrodes. The sodium chloride in water becomes ionized. When the power is switched on, a radial electric field is set up and the sodium and chloride ions start drifting in the influence of this field. The ring magnet below the cap produces a magnetic field in the vertical direction. The ions are moving in the horizontal plane of water. So their motion is perpendicular to the magnetic field. They experience a force which causes them to move in a circular path. This tendency of the ions to move in a circular path causes the whole water mass to rotate with it.

The magnetic Lorentz force $\vec{F}$ on a charged particles moving perpendicular to the magnetic field is given by $\vec{F}=q\;\vec{v}\times\vec{B}$, where $q$, $\vec{v}$ and $\vec{B}$ are the magnitudes of charge, velocity of the charge particle and the magnetic field, respectively. This force provides the centripetal force to make the charged particle move in circular path.

Problems from IIT JEE

Problem (IIT JEE 2013): A particle of mass $M$ and positive charge $Q$, moving with a constant velocity $\vec{u}_1={4\hat{\imath}}\;\mathrm{m/s}$, enters a region of uniform static magnetic field normal to the $x\text{-}y$ plane. The region of the magnetic field extends from $x=0$ to $x=L$ for all values of $y$. After passing through this region, the particle emerges on the other side after $10$ milliseconds with a velocity $\vec{u}_2={2\left(\sqrt{3}\hat{\imath}+\hat{\jmath}\right)}\;\mathrm{m/s}$. The correct statement(s) is (are)

  1. The direction of the magnetic field is $-z$ direction.
  2. The direction of the magnetic field is $+z$ direction.
  3. The magnitude of the magnetic field is $\frac{50\pi M}{3Q}$ units.
  4. The magnitude of the magnetic field is $\frac{100\pi M}{3Q}$ units.

Solution: The path followed by the particle is a circular arc PQ with $\vec{u}_1$ and $\vec{u}_2$ tangential at P and Q (see figure).


The $\vec{u}_2=2(\sqrt{3}\,\hat{\imath}+\hat{\jmath})$ gives $\theta=\pi/6$. For this motion, magnetic field ($\vec{B}$) shall be in $-z$ direction so that $Q\,\vec{v}\times\vec{B}$ provide necessary centripetal acceleration. From Newton's second law, \begin{align} QvB=\frac{Mv^2}{r}, \end{align} which gives \begin{align} \omega=\frac{v}{r}=\frac{QB}{M}, \end{align} where $v$ is velocity, $\omega$ is angular velocity and $r$ is radius of the circular path. The angular displacement in time $t$ is \begin{align} \theta=\omega t=\frac{QB}{M}t. \end{align} Substituting for $\theta$ and $t$, we get, \begin{align} B=\frac{50}{3}\frac{\pi M}{Q}. \end{align}


JEE Physics Solved Problems in Mechanics