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Moving Charges and Magnetism gives JEE Main two or three questions nearly every year, and they are mostly direct. The field of a wire, a loop, a solenoid and a toroid; the circular path of a charged particle; the conversion of a galvanometer. What trips students is mixing up which formula carries a 2π and which does not. Sort that once and this chapter pays for itself. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 2 to 3 questions every year, which is 8 to 12 marks.
- Fields of a wire, a loop and a solenoid are asked every year.
- The motion of a charged particle and meter conversions fill the rest.
What the syllabus says
In the syllabus (Unit 13, Magnetic Effects of Current and Magnetism): The Biot–Savart law and its application to a current-carrying circular loop, Ampere’s law applied to a long straight wire and a solenoid, force on a moving charge in uniform magnetic and electric fields, force on a current-carrying conductor in a magnetic field, the force between two parallel currents and the definition of the ampere, torque on a current loop in a uniform magnetic field, and the moving coil galvanometer with its sensitivity and its conversion to an ammeter and a voltmeter.
Taken out in 2024: the cyclotron.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Lorentz force | F = q(v × B), magnitude qvB sinθ |
| Long straight wire | B = μ0I/2πr |
| Centre of a circular coil | B = μ0NI/2r |
| Centre of a semicircular arc | B = μ0I/4r |
| Long solenoid | B = μ0nI |
| Toroid | B = μ0NI/2πr inside, zero outside |
| Radius and period | r = mv/qB, T = 2πm/qB |
| Radius from kinetic energy | r = √(2mK)/qB |
| Force between parallel wires | F/L = μ0I1I2/2πd |
| Magnetic moment of a loop | M = NIA |
| Galvanometer to voltmeter | R = V/Ig − G |
The form r = √(2mK)/qB is the one to use whenever a question gives kinetic energy rather than speed. It makes comparisons between a proton, a deuteron and an alpha particle a single substitution instead of two steps.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. A charged particle enters a uniform magnetic field at an angle of 30° to it. Its path is
- (a) a circle
- (b) a straight line
- (c) a helix
- (d) a parabola
Q2. The magnetic field at the centre of a semicircular arc of radius r carrying a current I is
- (a) μ0I/2r
- (b) μ0I/4r
- (c) μ0I/2πr
- (d) μ0I/4πr
Q3. A charged particle enters a uniform magnetic field with its velocity parallel to the field. Its path is
- (a) a circle
- (b) a helix
- (c) a straight line
- (d) a parabola
Q4. Two long parallel wires 1 m apart each carry a current of 1 A. The force per metre between them is
- (a) 2 × 10−7 N
- (b) 2 × 10−6 N
- (c) 10−7 N
- (d) 4π × 10−7 N
Q5. The magnetic field inside a toroid of N turns and mean radius r carrying current I is
- (a) μ0NI/2r
- (b) μ0NI/2πr
- (c) μ0NI
- (d) zero
Q6. The magnetic field at a point outside an ideal toroid is
- (a) μ0nI
- (b) zero
- (c) μ0I/2πr
- (d) infinite
Q7. A galvanometer of resistance G gives full scale deflection at current Ig. To convert it into a voltmeter of range V, the resistance to be joined in series is
- (a) V/Ig − G
- (b) V/Ig + G
- (c) IgG/V
- (d) G/V
Q8. The magnetic moment of a coil of N turns, area A, carrying current I is
- (a) NIA
- (b) IA/N
- (c) NI/A
- (d) NA/I
Q9. A proton and an alpha particle with the same kinetic energy enter the same magnetic field at right angles. The ratio of the radii of their paths is
- (a) 1 : 1
- (b) 1 : 2
- (c) 2 : 1
- (d) 1 : √2
Q10. The SI unit of magnetic flux is the
- (a) tesla
- (b) weber
- (c) henry
- (d) ampere
Answer key, with a hint for each
| Q1 | (c) | The component along the field is untouched and the component across it goes in a circle. Together they make a helix. |
| Q2 | (b) | Half of the full loop value μ0I/2r. |
| Q3 | (c) | The force is qv × B, which is zero when v is parallel to B. With no force acting, the particle simply carries straight on. A circle needs v perpendicular to B, and a helix needs a velocity that has both a parallel and a perpendicular part. |
| Q4 | (a) | μ0/2π = 2 × 10−7, and this is exactly how the ampere used to be defined. |
| Q5 | (b) | Apply Ampere’s law to a circle along the toroid, which encloses all N turns. |
| Q6 | (b) | No current is enclosed by a loop drawn outside, so the field there is zero. |
| Q7 | (a) | The series resistance must take the rest of the voltage, so R + G = V/Ig. |
| Q8 | (a) | M = NIA, and its unit is the ampere square metre. |
| Q9 | (a) | r = √(2mK)/qB. The alpha has 4 times the mass and 2 times the charge, and the two effects cancel exactly. |
| Q10 | (b) | One weber is one tesla square metre. The tesla is for the field itself. |
Now mark yourself honestly
JEE Main marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 25 Physics questions are out of 100.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in January.
The full chapter, on a clock
Ten questions tell you a little. Twenty-five tell you the truth.
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