Physics · Moving Charges And Magnetism · NEET
Two forces act on it. The electric force is F_E = qE (fixed, does not depend on speed). The magnetic force is F_B = qvB (grows with speed v). For zero deflection these must balance: qE = qvB. Cancel q from both sides and you get v = E/B. Only this one speed makes the two forces equal, so the charge goes straight.
It is v = E/B. Quick check with units: E is in volt/metre (or N/C) and B is in tesla. E/B gives metre/second, which is a speed. B/E would give the wrong units. So whenever you see 'undeflected' or 'zero deflection' in crossed fields, write v = E/B.
No. When you write qE = qvB, the charge q cancels out completely. Mass m never even appears (no acceleration term is needed at balance). So v = E/B is the same for an electron, a proton, or any ion. The selector picks a speed, not a specific particle.
If v is greater than E/B, the magnetic force qvB is bigger than qE, so the magnetic force wins and the charge bends. If v is less than E/B, the electric force qE is bigger, so it bends the other way. Both get deflected and are blocked by a slit. Only v = E/B passes straight through the slit.
Yes. E, B, and the velocity v are all mutually perpendicular. The magnetic force qv x B points opposite to the electric force qE only in this crossed arrangement. If E were parallel to B, the forces could not cancel and there would be no selection.
An electron (mass 9 x 10^-31 kg, charge 1.6 x 10^-19 C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9 x 10^-4 T perpendicular to its direction of motion. We wish to apply a uniform electric field E together with the magnetic field so that the electron does not deflect from its path. Then (c = 3 x 10^8 m/s):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It lets only charges of one chosen speed pass straight through, filtering out all other speeds. It is used at the entry of instruments like mass spectrometers, where you need a beam of particles all moving at the same known speed.
The selected speed is v = E/B, found by balancing the electric force qE against the magnetic force qvB. E is the electric field, B is the magnetic field, and they are perpendicular to each other and to v.
Yes. Since q cancels in qE = qvB, the condition v = E/B is the same regardless of the sign or size of the charge. For opposite charges both forces reverse direction together, so they still cancel at v = E/B.
The electric force is F = qE, which has no v in it. The magnetic force is F = qvB (from F = qv x B), which grows directly with speed. That is exactly why balancing them picks out one special speed.
