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Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 4 August 2021, Shift 1, Physics Q28

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

A potential difference of 800 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 5 mm and a uniform magnetic field is applied parallel to the plates horizontally. An electron projected vertically parallel to the plates, with a velocity of 2×1062\times10^6 m/s moves undeflected between the plates. The magnitude of the magnetic field will be ____ ×10−2\times 10^{-2} T.

Official answer

8

NTA final key.

Chapter
Moving Charges and Magnetism
Topic
Lorentz Force
Idea tested
Velocity Selector and Crossed Fields

Same idea in other shifts

Asked 4× in all
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  3. 26 Jul 2022, Shift 2 · Q47A velocity selector consists of electric field E⃗=Ek^\vec{E} = E\hat{k} and magnetic field B⃗=Bj^\vec{B} = B\hat{j} with B=12 mTB = 12\ mT. The value…MediumSingle 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 108 of the 174 JEE Main shifts held from 2017 to 2026.