Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas JEE Main2020PhysicsElectromagnetic InductionPractice

In the figure, a conducting rod of length l = 1 meter and mass m = 1 kg moves with an initial velocity, u = 5 m s - 1 . On a fixed horizontal frame containing inductor L = 2 H and resistance R = 1 Ω . PQ and MN are smooth, conducting wires. There is a uniform magnetic field of strength B = 1 T . Initially, there is no current in the inductor. Find the total charge in coulomb, flown through the inductor by the time ve

Correct answer

1

Step-by-step solution

Let i 1 and i 2 be the current through L and R at any time t ∴           i = i 1 + i 2   ⇒     B l v R = i 2 and B l v = L d i 1 d t Force on conducting rod = m d v d t = - i l B = - i 1 + B l v R l B ⇒         m d v =   - l B   i 1 d t - B 2 l 2 R v   d t   ⇒         m ∫ d v = - l B   ∫ i 1 d t - B 2 l 2 R   ∫ v   d t ⇒         m v f

Practice Electromagnetic Induction on Quantrex Academy →

More from Electromagnetic Induction

List-I contains four conducting loops lying in the XY plane, as shown in the figures. The loops are rotating about Z axis passing through the point O with time period T in clockwis 2026A 30 cm long solenoid has 10 turns per cm and area of 5 cm ^2 . The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is 10⁻⁵ V. Th 2026A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B = 0.4 (300t) Tesla. The normal to the plane of loop makes an angle of 60° with the field. 2026An inductor of inductance 10 mH having resistance of 100 , is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio 2026A metal rod of length L rotates about one end at origin with a uniform angular velocity . The magnetic field radially falls off as B(r) = B₀ e^ - r ; being a positive constant. The 2026A circular loop of radius 20 cm and resistance 2 is placed in a time varying magnetic field B = (2t^2 + 2t + 3) T . At t = 0 , for the plane of the loop being perpendicular to the 2026In the given circuit below inductance values of L₁ , L₂ and L₃ are same. The magnetic energy stored in the entire circuit is (U_t) and that stored in the L₂ inductor is (U_l) . U_t 2026A circular current loop of radius R is placed inside square loop of side length L(L >> R) such that they are co-planar and their centers coincide. The permeability of free space is 2026 Full Electromagnetic Induction list All NTA Abhyas JEE Main PYQs