The Question is from neet-physics electromagnetic-induction self-induction-and-mutal-induction, and it was created by the author Mr. ANIL KUMAR.G

Question

Two coaxial solenoids of radii $r_1$ and $r_2$ ($r_2 > r_1$) have the same length $l$ and the same number of turns $N$. If the current in the outer solenoid changes at a rate of $\frac{dI}{dt}$, what is the induced emf in the inner solenoid?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Faraday's Law of Induction and Magnetic Flux
 **Related Topics**: Solenoids, Mutual Inductance
 **Prerequisites**: Understanding of magnetic fields generated by solenoids, Faraday's Law ($\epsilon = -\frac{d\Phi}{dt}$), and magnetic flux ($\Phi = B \cdot A$).
## SOLUTION PATHWAY
**Given Information**:
  $r_1$ = radius of inner solenoid
  $r_2$ = radius of outer solenoid ($r_2 > r_1$)
  $l$ = length of both solenoids
  $N$ = number of turns in both solenoids
  $\frac{dI}{dt}$ = rate of change of current in the outer solenoid
**Method & Steps**:
 1. Find the magnetic field inside the outer solenoid.
  Reasoning: The magnetic field inside a solenoid is given by $B = \mu_0 n I$, where $n = \frac{N}{l}$ is the number of turns per unit length.
  Formula/Rule: $B_{outer} = \mu_0 \frac{N}{l} I$
 2. Calculate the magnetic flux through the inner solenoid due to the magnetic field of the outer solenoid.
  Working: The magnetic field inside the outer solenoid is uniform and is equal to $B_{outer}$. The area of the inner solenoid is $A_1 = \pi r_1^2$.  The magnetic flux through the inner solenoid is given by $\Phi = B_{outer} A_1 N$.
  Key Point: Since the inner solenoid is inside the outer solenoid, the magnetic field of the outer solenoid passes through the inner solenoid. 
  Formula/Rule: $\Phi = \mu_0 \frac{N}{l} I \pi r_1^2 N = \mu_0 \frac{N^2}{l} I \pi r_1^2$
 3. Apply Faraday's Law to find the induced emf in the inner solenoid.
  Working: The induced emf is the rate of change of magnetic flux.  $\epsilon = -\frac{d\Phi}{dt} = -\frac{d}{dt} (\mu_0 \frac{N^2}{l} I \pi r_1^2) = -\mu_0 \frac{N^2}{l} \pi r_1^2 \frac{dI}{dt}$ . The negative sign indicates the direction of the induced emf (Lenz's Law), but the magnitude is what's asked for.
  Result: $\epsilon = \mu_0 \frac{N^2}{l} \pi r_1^2 \frac{dI}{dt}$
  Verification: The units are Volts, which is consistent with induced emf.
 ## OPTION ANALYSIS
A] 0 :x: Incorrect.  A changing magnetic flux induces an emf.
B] ✓: Correct. This matches the derived expression for induced emf.
C] :x: Incorrect. This uses the radius of the outer solenoid, not the inner.
D] :x: Incorrect. This formula is not applicable to the situation of a solenoid within another solenoid. 
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the number of turns in the inner solenoid when calculating the flux: Remember that each turn of the inner solenoid experiences the magnetic flux, so it must be multiplied by N.
  2.Using the wrong area: The area relevant is the area of the inner solenoid, not the outer.
 **Quick Tips**:
  1.Remember the formula for the magnetic field inside a solenoid and Faraday's law of induction.
  2.Focus on the magnetic flux through the inner solenoid only.

Mr. ANIL KUMAR.G · Answer

The correct Option is "0"

Mr. ANIL KUMAR.G · Answer

$\mu_0 N^2 \pi r_1^2 \frac{dI}{dt} \frac{1}{l}$

Mr. ANIL KUMAR.G · Answer

$\mu_0 N^2 \pi r_2^2 \frac{dI}{dt} \frac{1}{l}$

Mr. ANIL KUMAR.G · Answer

$\mu_0 N^2 \pi (r_2^2 - r_1^2) \frac{dI}{dt} \frac{1}{l}$

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