A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Omega then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ____ N.
Motional EMF diagram for Q41 - JEE Main 2026 Morning
A conducting rod AB moves on a U-shaped rail in a perpendicular magnetic field.

Solution & Explanation

### Related Formula E = B l v i = fracER F_B = i l B = fracB^2 l^2 vR ### Core Logic To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current. F_textext = F_B ### Step 1: Calculate External Force Given values: B = 0.10text T l = 1text m v = 1.5text m/s R = 2\, Omega F_textext = fracB^2 l^2 vR F_textext = frac(0.10)^2 times (1)^2 times 1.52 F_textext = frac0.01 times 1.52 = frac0.0152 = 0.0075 F_textext = 7.5 times 10^-3text N
Motional EMF solution diagram for Q41 - JEE Main 2026 Morning
A conducting rod AB moves on a U-shaped rail in a perpendicular magnetic field.
### Pattern Recognition Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always F = fracB^2 L^2 vR. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

More Electromagnetic Induction Previous-Year Questions — Page 4

Q53 jee_main_2024_31_jan_evening Faraday's Law
The magnetic flux phi (in weber) linked with a closed circuit of resistance 8 \, Omega varies with time (in seconds) as phi = 5t^2 - 36t + 1. The induced current in the circuit at t = 2 text s is ________ A.
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula varepsilon = -fracdphidt I = frac|varepsilon|R ### Core Logic Calculate the time derivative of the magnetic flux to find the induced EMF. Evaluate it at the requested time, and then apply Ohm's law to find the current magnitude. ### Step 1: Calculate Induced EMF varepsilon = -fracddt (5t^2 - 36t + 1) varepsilon = -(10t - 36) At t = 2 text s: varepsilon = - (10 times 2 - 36) varepsilon = -(20 - 36) = 16 text V ### Step 2: Calculate Induced Current I = fracvarepsilonR = frac168 = 2 text A ### Pattern Recognition Flux polynomials (At^2 - Bt + C) instantly trigger a simple derivative test. Remember to drop the negative sign for final current magnitude unless direction is specifically asked. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction
Q44 jee_main_2024_31_jan_morning Faraday's Law
A coil is placed perpendicular to a magnetic field of 5000 mathrm~T. When the field is changed to 3000 mathrm~T in 2mathrms, an induced emf of 22 mathrm~V is produced in the coil. If the diameter of the coil is 0.02 mathrm~m, then the number of turns in the coil is:
  • A. 7
  • B. 70
  • C. 35
  • D. 140

Solution

### Related Formula varepsilon = N left| fracDeltaphiDelta t right| Deltaphi = (Delta B) A costheta ### Core Logic Given data: Initial Magnetic Field, B_i = 5000mathrm\,T Final Magnetic Field, B_f = 3000mathrm\,T Time interval, Delta t = 2mathrm\,s Diameter, d = 0.02mathrm\,m Rightarrow r = 0.01mathrm\,m Induced emf, varepsilon = 22mathrm\,V Change in magnetic field magnitude |Delta B| = 5000 - 3000 = 2000mathrm\,T. Area of the coil A = pi r^2 = pi (0.01)^2 = 10^-4pi mathrm\,m^2. ### Step 2: Equation Evaluation Deltaphi = |Delta B| A = (2000) pi (0.01)^2 = 0.2pi Using Faraday's Law: 22 = N left( frac0.2pi2 right) 22 = N (0.1pi) Taking pi approx 22/7: 22 = N left( 0.1 times frac227 right) 1 = fracN70 Rightarrow N = 70 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction
Q57 jee_main_2024_31_jan_morning Mutual Inductance
A small square loop of wire of side ell is placed inside a large square loop of wire of side L (L = ell^2). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is sqrtx times 10^-7mathrm\ H, where x =
Numerical Answer. Answer: 128 to 128

Solution

### Related Formula M = fracphi_2i_1 B_textstraight wire segment = fracmu_0 i4pi d (sintheta_1 + sintheta_2) ### Core Logic
Mutual Inductance diagram for Q57 - JEE Main 2024 Morning
Mutual Inductance diagram for Q57 - JEE Main 2024 Morning
Assume a current i flows through the larger square loop of side L. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side ell. The magnetic field at the center of the large square loop (distance d = L/2 from each side, angles 45^circ): B = 4 times left[ fracmu_0 i4pi (L/2) (sin 45^circ + sin 45^circ) right] B = fracmu_0 ipi (L/2) left( frac2sqrt2 right) B = frac2sqrt2 mu_0 ipi L ### Step 2: Mutual Inductance Calculation Flux linkage for the inner loop: phi = B cdot ell^2 phi = frac2sqrt2 mu_0 ipi L ell^2 Given L = ell^2: phi = frac2sqrt2 mu_0 ipi (ell^2) ell^2 = frac2sqrt2 mu_0 ipi Mutual inductance M: M = fracphii = frac2sqrt2 mu_0pi Using mu_0 = 4pi times 10^-7: M = frac2sqrt2 times 4pi times 10^-7pi M = 8sqrt2 times 10^-7mathrm\,H M = sqrt128 times 10^-7mathrm\,H Comparing with sqrtx times 10^-7, we get x = 128. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

More Electromagnetic Induction Questions — jee_main_2026_21_jan_morning

Practice all Electromagnetic Induction previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...