JEE Main 2024 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 1 February 2024, Shift 2, Physics Q58

Question 58 of 90 in this shift, Physics question 28 of 30, Section B.

A coil of 200 turns and area

0.20\ \mathrm{m^2}

is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is

\frac{2\pi}{\beta}

volt. The value of

\beta

is ________.

Official answer

NTA final key.

Chapter: Electromagnetic Induction
Topic: Faraday's Law
Idea tested: AC Generator Principle

Same idea in other shifts

Asked 5× in all

24 Jun 2022, Shift 2 · Q58A circular coil of 1000 turns each with area 1m $^2$ is rotated about its vertical diameter at the rate of one revolution per second in a…EasyNumerical value
26 Aug 2021, Shift 2 · Q28A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s $^{-1}$ in a uniform…EasyNumerical value
2 Sep 2020, Shift 1 · Q24A circular coil of radius 10 cm is placed in a uniform magnetic field of $3.0\times10^{-5}$ T with its plane perpendicular to the field…MediumNumerical value
3 Sep 2020, Shift 1 · Q14An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in a magnetic field as shown in the figure.…MediumSingle correctHas a figure

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 108 of the 174 JEE Main shifts held from 2017 to 2026.