JEE Main 2025 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 2 April 2025, Shift 1, Physics Q38

Question 38 of 75 in this shift, Physics question 13 of 25, Section A.

Let

B_1

be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let

B_2

be the magnitude of magnetic field at an axial distance '

x

' from the center. For

x:R=3:4

\frac{B_2}{B_1}

is :

(1)25 : 16
(2)16 : 25
(3)4 : 5
(4)64 : 125Official answer

Official answer

Option 4

NTA final key.

Chapter: Moving Charges and Magnetism
Topic: Biot-Savart Law
Idea tested: Ratio of Axial to Central Magnetic Field

Same idea in other shifts

Asked 3× in all

8 Apr 2023, Shift 2 · Q59The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre…MediumNumerical value
24 Jun 2022, Shift 1 · Q49The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point…EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.