The Question is from neet-physics electromagnetic-induction , and it was created by the author SUBHANI

Question

A toroidal solenoid with $N$ turns and mean radius $r$ carries a current $I$. A single loop of wire is wrapped around the toroid.  If the current in the solenoid is changing at a rate $\frac{dI}{dt}$, what is the induced emf in the single loop?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: 3
## CONCEPT FRAMEWORK
 **Primary Concept**: Faraday's Law of Induction
 **Related Topics**: Magnetic flux, toroidal solenoid, induced EMF
 **Prerequisites**: Understanding of magnetic fields generated by solenoids, Faraday's Law, and calculus.
## SOLUTION PATHWAY
**Given Information**:
  $N$ = number of turns in the toroidal solenoid
  $r$ = mean radius of the toroid
  $I$ = current in the solenoid
  $\frac{dI}{dt}$ = rate of change of current
  A single loop of wire is wrapped around the toroid.
  [Important conditions] The toroid is assumed to be ideal (no leakage flux).
**Method & Steps**:
 1. Find the magnetic field inside the toroid.
  Reasoning: The magnetic field inside a toroid is given by $B = \frac{\mu_0 N I}{2\pi r}$.
  Formula/Rule: $B = \frac{\mu_0 N I}{2\pi r}$
 2. Calculate the magnetic flux through the single loop.
  Working: The magnetic flux through the single loop is given by $\Phi = B A$, where A is the area of the loop. The area of the loop is not explicitly given, but it's implied that the loop encircles the toroid.  Since the area of the loop is not specified, we represent it as A. Thus, $\Phi = \frac{\mu_0 N I A}{2\pi r}$.
  Key Point: The magnetic field is uniform across the cross-section of the toroid.
 3. Apply Faraday's Law to find the induced EMF.
  Reasoning: Faraday's Law states that the induced EMF is equal to the negative rate of change of magnetic flux: $\mathcal{E} = -\frac{d\Phi}{dt}$.
  Working: $\mathcal{E} = -\frac{d}{dt}(\frac{\mu_0 N I A}{2\pi r}) = -\frac{\mu_0 N A}{2\pi r} \frac{dI}{dt}$. The negative sign indicates the direction of the induced EMF (Lenz's Law), but the magnitude is what we're interested in. Therefore, the magnitude of the induced emf is $\frac{\mu_0 N A}{2\pi r} \frac{dI}{dt}$.
  Result: $\mathcal{E} = \frac{\mu_0 N A}{2\pi r} \frac{dI}{dt}$
  Verification: The units are consistent with EMF (Volts).
 ## OPTION ANALYSIS
[A] 0 :x: This would only be true if the magnetic flux were constant, which it isn't since the current is changing.
[B] $\mu_0 N A \frac{dI}{dt}$ :x: This formula is missing the $2\pi r$ term in the denominator, which accounts for the geometry of the toroid.
[C] $\mu_0 \frac{N}{2\pi r} A \frac{dI}{dt}$ ✓: This is the correct formula, derived from Faraday's Law and the magnetic field inside a toroid.
[D] $\frac{\mu_0}{2\pi r} A \frac{dI}{dt}$ :x: This formula is missing the $N$ term, which represents the number of turns in the solenoid and is crucial for determining the magnetic field strength.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Forgetting the $2\pi r$ term in the denominator when calculating the magnetic field inside a toroid.
  2. Incorrectly applying Faraday's Law; forgetting to take the derivative with respect to time or neglecting the area term.
 **Quick Tips**:
  1. Remember the formula for the magnetic field inside a toroid. 
  2. Always clearly identify the area through which the magnetic flux passes.

SUBHANI · Answer

0

SUBHANI · Answer

$\mu_0 N A \frac{dI}{dt}$

SUBHANI · Answer

The correct Option is "$\mu_0 \frac{N}{2\pi r} A \frac{dI}{dt}$"

SUBHANI · Answer

$\frac{\mu_0}{2\pi r} A \frac{dI}{dt}$

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