A 5 mg particle carrying a charge of 5pi x 10^-6 C is moving with velocity of (3i + 2k) x 10^-2 m/s in a region having magnetic field B = 0.1 k Wb/m^2. It moves a distance of alpha meter along k when it completes 5 revolutions. The value of alpha is _______.

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q48 – JEE-MAIN 2026

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Home Physics Magnetism and MatterQuestion #240

5 mg

particle carrying a charge of

5 π \times 10^{- 6} C

is moving with velocity of

(3 \hat{i} + 2 \hat{k}) \times 10^{- 2} m/s

in a region having magnetic field

\vec{B} = 0.1 \hat{k} {Wb/m}^{2}

. It moves a distance of

α

meter along

\hat{k}

when it completes 5 revolutions. The value of

α

is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Helical Motion

When a charged particle moves in a uniform magnetic field with a velocity component parallel to the field and a component perpendicular to the field, its path is a helix.

🧠 Full Solution Path

Step 1: Identify Velocity Components and Convert Units✦ Active

The mass of the particle is $m = 5 mg = 5 \times 10^{- 6} kg$ . The charge is $q = 5 π \times 10^{- 6} C$ . The velocity vector is $\vec{v} = (3 \hat{i} + 2 \hat{k}) \times 10^{- 2} m/s$ . The magnetic field is $\vec{B} = 0.1 \hat{k} T$ . We need to identify the components of velocity parallel and perpendicular to the magnetic field.

v_{∥} = 2 \times 10^{- 2} m/s (along \hat{k})

v_{⊥} = 3 \times 10^{- 2} m/s (along \hat{i})

Step 2: Calculate the Time Period of One Revolution○ Expand

The time period for one complete revolution of the circular motion (due to $v_{⊥}$ ) in the magnetic field is given by:

T = \frac{2 π m}{q B}

T = \frac{2 π (5 \times 10^{- 6} kg)}{(5 π \times 10^{- 6} C) (0.1 T)} = \frac{10 π \times 10^{- 6}}{0.5 π \times 10^{- 6}} = \frac{10}{0.5} = 20 s

💡 Teacher's Secret Hint

Ensure all units are in SI before calculation.

Step 3: Calculate the Total Distance Moved Along the Magnetic Field○ Expand

The particle completes 5 revolutions. The total time taken for 5 revolutions is $t_{t o t a l} = 5 T$ . During this time, the particle moves along the direction of the magnetic field (along $\hat{k}$ ) due to its parallel velocity component $v_{∥}$ . The distance moved, $α$ , is:

α = v_{∥} \times t_{t o t a l} = v_{∥} \times (5 T)

α = (2 \times 10^{- 2} m/s) \times (5 \times 20 s)

α = (2 \times 10^{- 2} m/s) \times (100 s)

α = 2 m

💡 Teacher's Secret Hint

Remember that the parallel velocity component is unaffected by the magnetic field and contributes to the linear displacement.

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Physics Question 48 – JEE-MAIN 2026

Clue 1 of 3: Helical Motion

🧠 Full Solution Path