PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 4 · JEE Main

Moving Charges and Magnetism

Chapter-wise practice and revision for JEE Main: concise notes on the Lorentz force and charged-particle motion, the Biot–Savart and Ampère laws, forces and torques on currents, and the galvanometer, concept figures, and 50 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

50 selected questions6 topics5 concept figuresPreview · first 3 of 50 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 213 previous-year JEE questions in the PiTheory bank

Lorentz force59
Biot–Savart52
Galvanometer45
Torque22
Force on currents18
Ampère's law17

Contents

  1. 1Force on a moving charge and its motionQ1–Q12
  2. 2Biot–Savart law: fields of wires and loopsQ13–Q23
  3. 3Ampère's law, thick wires and solenoidsQ24–Q28
  4. 4Force on currents and between wiresQ29–Q33
  5. 5Magnetic moment and torque on a loopQ34–Q40
  6. 6Moving coil galvanometer, ammeter and voltmeterQ41–Q50
  7. ✓Last-day checklist
01

Topic 1 · Q1–Q12

Force on a moving charge and its motion

59 of the 213 JEE PYQs in the bank test this as their main idea

Look first

vFB out of page: r = mv/qB
The force is always towards the centre.
Bpitchpitch = v cosθ · 2πm/qB
Parallel part moves on; perpendicular part circles.

Quick notes

$$\vec F=q(\vec E+\vec v\times\vec B),\qquad r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB},\qquad T=\frac{2\pi m}{qB},\qquad f=\frac{qB}{2\pi m}$$

The magnetic force is perpendicular to $\vec v$ and $\vec B$: it does no work, so speed and kinetic energy stay constant. Period does not depend on speed (cyclotron principle).

SituationResult
$\vec v\parallel\vec B$no force, straight line
$\vec v\perp\vec B$circle
At angle $\theta$helix: $r=\dfrac{mv\sin\theta}{qB}$, pitch $=v\cos\theta\cdot T$
Crossed $E$ and $B$, undeflected$v=\dfrac EB$ (velocity selector)
Accelerated through $V$ first$r=\dfrac1B\sqrt{\dfrac{2mV}{q}}$
Same $K$: p, d, $\alpha$$r\propto\sqrt m/q$: $1:\sqrt2:1$
  • $\vec a\perp\vec B$ always: $\vec a\cdot\vec B=0$.
  • Electron: $\vec F=-e\,\vec v\times\vec B$ (reverse the direction).
  • Along a solenoid's axis $\vec v\parallel\vec B$: no force.
  • Field leaving region of width $d$: deflection angle $\sin\theta=\dfrac dr$.
Traps. Order of the cross product. Use the perpendicular component of $v$ for the radius.

Practice: first 3 of 12 questions

Q1JEE Main 2026 (2 Apr, Shift 1)Easy

$1\ \mu$C charge moving with velocity $\vec v=(\hat i-2\hat j+3\hat k)$ m/s in the region of magnetic field $\vec B=(2\hat i+3\hat j-5\hat k)$ T. The magnitude of force acting on it is $\sqrt\alpha\times10^{-6}$ N. The value of $\alpha$ is ______.

Numerical answer type

Answer and solution

Answer: 171

Concept. Magnetic force $\vec F=q\,\vec v\times\vec B$. Work out the cross product with the determinant.

Step 1. Cross product. $(\hat i-2\hat j+3\hat k)\times(2\hat i+3\hat j-5\hat k)=(10-9)\hat i+(6+5)\hat j+(3+4)\hat k=\hat i+11\hat j+7\hat k$.

Step 2. Magnitude. $|\vec F|=10^{-6}\sqrt{1+121+49}=\sqrt{171}\times10^{-6}$ N.

Answer. $\alpha=171$.

Point to note. $q\,\vec v\times\vec B$ by determinant.

Q2JEE Main 2026 (2 Apr, Shift 2)Medium

A particle having charge $10^{-9}$ C moving in $x$-$y$ plane in fields of $0.4\hat j$ N/C and $4\times10^{-3}\hat k$ T experiences a force of $(4\hat i+2\hat j)\times10^{-10}$ N. The velocity of the particle at that instant is ______ m/s.

  1. A$50\hat i+100\hat j$
  2. B$100\hat i+50\hat j$
  3. C$-50\hat i+100\hat j$
  4. D$50\hat i-100\hat j$
Answer and solution

Answer: Option A: $50\hat i+100\hat j$

Concept. Lorentz force $\vec F=q(\vec E+\vec v\times\vec B)$. With $\vec v=v_x\hat i+v_y\hat j$ and $\vec B=B\hat k$: $\vec v\times\vec B=v_yB\,\hat i-v_xB\,\hat j$.

Step 1. Components. $x$: $qv_yB=4\times10^{-10}$; $y$: $qE-qv_xB=2\times10^{-10}$.

Step 2. Solve. $qB=4\times10^{-12}$: $v_y=100$; $4\times10^{-10}-4\times10^{-12}v_x=2\times10^{-10}\Rightarrow v_x=50$.

Answer. $50\hat i+100\hat j$ (Option A).

Point to note. $q(\vec E+\vec v\times\vec B)$ componentwise.

Q3JEE Main 2026 (5 Apr, Shift 1)Medium

The charged particle moving in a uniform magnetic field of $(3\hat{i} + 2\hat{j})$ T has an acceleration $\left(4\hat{i} - \dfrac{x}{2}\hat{j}\right)\ \text{m/s}^2$. The value of $x$ is

Numerical answer type

Answer and solution

Answer: 12

Concept. The magnetic force (and hence the acceleration) is always perpendicular to $\vec B$: $\vec a\cdot\vec B=0$.

Step 1. Dot product. $4\times3-\dfrac x2\times2=0$.

Answer. $x=12$.

Point to note. $\vec a\cdot\vec B=0$.

47 more questions and 5 more topics in the full chapter

  1. 2Biot–Savart law: fields of wires and loops11 questions with solutions
  2. 3Ampère's law, thick wires and solenoids5 questions with solutions
  3. 4Force on currents and between wires5 questions with solutions
  4. 5Magnetic moment and torque on a loop7 questions with solutions
  5. 6Moving coil galvanometer, ammeter and voltmeter10 questions with solutions
  6. ✓Last-day checklist

The complete Moving Charges and Magnetism chapter is in the Class 12 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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