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

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q29 – JEE-MAIN 2026

← Back

Home Physics Moving ChargesQuestion #344

A particle having charge

10^{- 9}

C moving in x-y plane in fields of

0.4 \hat{j}

N/C and

4 \times 10^{- 3} \hat{k}

T experiences a force of

(4 \hat{i} + 2 \hat{j}) \times 10^{- 10}

N. The velocity of the particle at that instant is _______ m/s.

Clue 1 of 3: Lorentz Force

The total force on a charged particle moving in combined electric and magnetic fields is given by the Lorentz force law.

🧠 Full Solution Path

Step 1: Apply Lorentz Force Law✦ Active

The total force $\vec{F}$ on a charged particle $q$ moving with velocity $\vec{v}$ in an electric field $\vec{E}$ and a magnetic field $\vec{B}$ is given by the Lorentz force law:

\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})

Given values are $q = 10^{- 9}$ C, $\vec{E} = 0.4 \hat{j}$ N/C, $\vec{B} = 4 \times 10^{- 3} \hat{k}$ T, and $\vec{F} = (4 \hat{i} + 2 \hat{j}) \times 10^{- 10}$ N. Let the velocity be $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j}$ since the particle moves in the x-y plane. Substitute these into the Lorentz force equation:

(4 \hat{i} + 2 \hat{j}) \times 10^{- 10} = 10^{- 9} (0.4 \hat{j} + (v_{x} \hat{i} + v_{y} \hat{j}) \times (4 \times 10^{- 3} \hat{k}))

Step 2: Simplify and Calculate Magnetic Force Term○ Expand

Divide the entire equation by $10^{- 9}$ and rearrange:

0.4 \hat{i} + 0.2 \hat{j} = 0.4 \hat{j} + (v_{x} \hat{i} + v_{y} \hat{j}) \times (4 \times 10^{- 3} \hat{k})

Now, calculate the cross product term $\vec{v} \times \vec{B}$ :

(v_{x} \hat{i} + v_{y} \hat{j}) \times (4 \times 10^{- 3} \hat{k}) = (v_{x}) (4 \times 10^{- 3}) (\hat{i} \times \hat{k}) + (v_{y}) (4 \times 10^{- 3}) (\hat{j} \times \hat{k})

Using $\hat{i} \times \hat{k} = - \hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$ :

= (4 \times 10^{- 3} v_{x}) (- \hat{j}) + (4 \times 10^{- 3} v_{y}) (\hat{i})

= (4 \times 10^{- 3} v_{y}) \hat{i} - (4 \times 10^{- 3} v_{x}) \hat{j}

💡 Teacher's Secret Hint

Remember the cyclic properties of unit vector cross products: $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ .

Step 3: Equate Components and Solve for Velocity○ Expand

Substitute the magnetic force term back into the main equation:

0.4 \hat{i} + 0.2 \hat{j} = 0.4 \hat{j} + (4 \times 10^{- 3} v_{y}) \hat{i} - (4 \times 10^{- 3} v_{x}) \hat{j}

Rearrange to group terms:

0.4 \hat{i} + (0.2 - 0.4) \hat{j} = (4 \times 10^{- 3} v_{y}) \hat{i} - (4 \times 10^{- 3} v_{x}) \hat{j}

0.4 \hat{i} - 0.2 \hat{j} = (4 \times 10^{- 3} v_{y}) \hat{i} - (4 \times 10^{- 3} v_{x}) \hat{j}

Equate the coefficients of $\hat{i}$ and $\hat{j}$ components:

For $\hat{i}$ component:

0.4 = 4 \times 10^{- 3} v_{y} ⟹ v_{y} = \frac{0.4}{4 \times 10^{- 3}} = \frac{4 \times 10^{- 1}}{4 \times 10^{- 3}} = 10^{2} = 100 m/s

For $\hat{j}$ component:

- 0.2 = - 4 \times 10^{- 3} v_{x} ⟹ v_{x} = \frac{0.2}{4 \times 10^{- 3}} = \frac{2 \times 10^{- 1}}{4 \times 10^{- 3}} = \frac{1}{2} \times 10^{2} = 50 m/s

Thus, the velocity of the particle is $\vec{v} = 50 \hat{i} + 100 \hat{j}$ m/s.

💡 Teacher's Secret Hint

Carefully handle the signs and powers of 10 during calculations.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 29 – JEE-MAIN 2026

Clue 1 of 3: Lorentz Force

🧠 Full Solution Path