The diagram shows a solenoid carrying a time-varying current, I=I0t . On the axis of this solenoid, a ring has been placed as shown in the figure at the right. The mutual inductance of the ring and the solenoid is M and the self-inductance of the ring is L . If the resistance of the ring is R , then maximum current which can flow through the ring is,

Question

The diagram shows a solenoid carrying a time-varying current, I=I0t . On the axis of this solenoid, a ring has been placed as shown in the figure at the right. The mutual inductance of the ring and the solenoid is M and the self-inductance of the ring is L . If the resistance of the ring is R , then maximum current which can flow through the ring is, <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-zopoaihs3atkrnja.jpg" /> (A) ((. 2 M + L .) I0/R) (B) (M I0/R) (C) ((. 2 M - L .) I0/A) (D) ((. M + L .) I0/R)

Tardigrade Admin · Accepted Answer

Magnetic flux linking due to mutual inductance is φ=MI=MI0t Hence, induced emf |e|=(d φ/d t)=MI0 Current through ring, i=(e/R)=(M I0/R) .

$\frac{( 2 M + L ) I _{0}}{R}$

$\frac{M I _{0}}{R}$

$\frac{( 2 M - L ) I _{0}}{A}$

$\frac{( M + L ) I _{0}}{R}$

Magnetic flux linking due to mutual inductance is $ϕ = M I = M I_{0} t$
Hence, induced emf $∣ e ∣ = \frac{d ϕ}{d t} = M I_{0}$
Current through ring, $i = \frac{e}{R} = \frac{M I _{0}}{R}$ .

R(2M+L)I0​​

RMI0​​

A(2M−L)I0​​

R(M+L)I0​​

Magnetic flux linking due to mutual inductance is ϕ=MI=MI0​t Hence, induced emf ∣e∣=dtdϕ​=MI0​ Current through ring, i=Re​=RMI0​​ .

$\frac{( 2 M + L ) I _{0}}{R}$

$\frac{M I _{0}}{R}$

$\frac{( 2 M - L ) I _{0}}{A}$

$\frac{( M + L ) I _{0}}{R}$

Magnetic flux linking due to mutual inductance is $ϕ = M I = M I_{0} t$
Hence, induced emf $∣ e ∣ = \frac{d ϕ}{d t} = M I_{0}$
Current through ring, $i = \frac{e}{R} = \frac{M I _{0}}{R}$ .