Consider I₁ and I₂ are the currents flowing simultaneously in two nearby coils 1 2, respectively. If L₁ = self inductance of coil 1, M₁₂ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be

(C) ε₁ = -L₁dI₁/dt - M₁₂dI₂/dt

(A) ε₁ = -L₁dI₁/dt + M₁₂dI₂/dt

(B) ε₁ = -L₁dI₁/dt - M₁₂dI₁/dt

(D) ε₁ = -L₁dI₂/dt - M₁₂dI₁/dt

Electromagnetic Induction and Alternating Currents: JEE Main 2025 Physics question

JEE Main 2025 · 29 Jan, Morning shift · Q26

JEE Main 202529 Jan, Morning shiftPhysicsMediumElectromagnetic Induction and Alternating Currents149× since 2002

Consider

I_1

and

I_2

are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If

L_1

= self inductance of coil 1,

M_{12}

= mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be

JEE Main 2025 Electromagnetic Induction and Alternating Currents Question with Solution (29 Jan Morning shift, Q26) | JEE Arc