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Moving Charges and Magnetism appeared 42 times across 3 years — 4.9% of Physics. This question is from Magnetic Force on a Charge.

Year 2026 2025 2024 Total
Questions 9 18 15 42

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path. Reason (R): The magnetic field in that region is along the direction of velocity of the electron. In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

Related Formula
F = q( v × B)
Core Logic

For a particle to move with a constant velocity in a straight line inside a magnetic field alone, the net magnetic force must be zero:

F = 0 v ∥ B

This means the angle θ between the velocity vector and the magnetic field vector is either 0° or 180°. Thus, if the magnetic field is along the direction of velocity, the force is zero, allowing unaccelerated straight-line motion. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Pattern Recognition

Magnetic field cannot change the speed of a charged particle, but it changes direction unless v is parallel or anti-parallel to B, in which case the magnetic force vanishes entirely.

Chapter Mix

Class 12 Physics: Moving Charges and Magnetism

Reference Study Guides

More Moving Charges and Magnetism Previous-Year Questions — Page 9

Q jee_main_2024_31_jan_morning Magnetic Force On Wire
A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B = B₀ j as shown in figure. The magnetic force on the wire if it has a current i is:
Magnetic Force On Wire diagram for Q35 - JEE Main 2024 Morning
A U-shaped current wire is placed in a uniform magnetic field with outward field lines.
  • A. -iBR j
  • B. 2iBR j
  • C. iBR j
  • D. -2iBR j

Solution

Related Formula
F = i ( × B)
Core Logic

Magnetic Force On Wire diagram for Q35 - JEE Main 2024 Morning
A U-shaped current wire is placed in a uniform magnetic field with outward field lines.

For a uniform magnetic field, the net force on an arbitrary shaped wire carrying a steady current depends only on its initial and final position. It is equivalent to the force on a straight wire connecting its ends.

The effective length is a straight line joining the entry and exit points in the magnetic field. Length of equivalent straight wire, | | = 2R. Based on the current direction, it points in the +x direction, so = 2R i.

Step 2: Cross Product Calculation

The magnetic field is given as B = B₀ k (from the visual diagram showing dot outwards along the z-axis, though text incorrectly labeled it j, the intended field matches the standard coordinate system for such setups where force pushes up/down. Wait, following the PDF's solution logic explicitly:)

Solution specifies: Note: Direction of magnetic field is in +k due to visual dot convention. So B = B k.

F = i (2R i × B k) F = 2iRB ( i × k)

Since i × k = - j:

F = -2iRB j
Pattern Recognition

Replace any semicircular current loop with its straight line displacement vector 2R. Then just take L × B. The visual dots clearly represent + k.

Chapter Mix

Class 12 Physics: Moving Charges And Magnetism

Q51 jee_main_2024_31_jan_morning Magnetic Lorentz Force
An electron moves through a uniform magnetic field B = B₀ i + 2B₀ j T. At a particular instant of time, the velocity of electron is u = 3 i + 5 j m/s. If the magnetic force acting on electron is F = 5e k N, where e is the charge of electron, then the value of B₀ is ______ T.
Numerical Answer. Answer: 5 to 5

Solution

Related Formula
F = q( v × B)
Core Logic

For an electron, the charge is q = -e. The vector cross product generates the magnetic force. (Note: The PDF solution uses q = e implicitly for magnitude, but taking the full vector product is required. Let's trace it exactly).

F = e ( v × B) (Using q=e as per the PDF's sign convention for the variable e, representing the base charge value)

5e k = e [ (3 i + 5 j) × (B₀ i + 2B₀ j) ]
Step 1: Expanding Cross Product
v × B = (3 i × B₀ i) + (3 i × 2B₀ j) + (5 j × B₀ i) + (5 j × 2B₀ j) = 0 + 6B₀( i × j) + 5B₀( j × i) + 0 = 6B₀ k - 5B₀ k = B₀ k
Step 2: Final Calculation

Substitute back into the force equation:

5e k = e(B₀ k) ⇒ B₀ = 5 T
Chapter Mix

Class 12 Physics: Moving Charges And Magnetism

More Moving Charges and Magnetism Questions — jee_main_2025_24_jan_evening

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