A thin conducting rod MN of mass 20 gm, length 25 cm and resistance 10 Ω.. | JEE Advanced 2023

Like and subscribe if you enjoyed :).
Q: A thin conducting rod MN of mass 20 gm, length 25 cm and resistance 10 Ω is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field 𝐵0 = 4 T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time 𝑡 = 0 and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.
[Given: The acceleration due to gravity 𝑔 = 10 m s−2 and 𝑒−1 = 0.4]
List 1
(P) At 𝑡 = 0.2 s, the magnitude of the induced emf in Volt
(Q) At 𝑡 = 0.2 s, the magnitude of the magnetic force in Newton
(R) At 𝑡 = 0.2 s, the power dissipated as heat in Watt
(S) The magnitude of terminal velocity of the rod in m s−1
List 2
(1) 0.07
(2) 0.14
(3) 1.20
(4) 0.12
(5) 2.00

[effortlessones....]

btw very good explanation really helped me

[SaurabhAbhishek-ppru]

Thanks! For the video but you have taken wrong polarity of battery! But the dirn of force due to magnetic field is correct 1:57

[elamaranyuvaraj]

sir i have a doubt in 4:30, from the equation mg-ilb=ma you have substituted i in terms of v/r right, but here this v denotes the potenial difference between the two ends of the rod, how can assume that v as the velocity of the rod...using ohms we can only derive a reln between current and potenial difference not velocity...am i right?

[effortlessones....]

V cross b for polarity sir apne ulta likha hai

[AdityaSingh-ijph]

Thank god my paranoid a&& had the same thought process during the exam and did the same procedure. Hence got it correct eventually💀💀