The Question is from neet-physics electromagnetic-induction induced-electric-field-eddy-currents, and it was created by the author VENKATESWARARAO

Question

A conducting disc rotates with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of the disc.  A small conducting ring is placed on the rim of the disc.  If $r$ is the radius of the disc, then the induced emf between the center and the ring is:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [A]
## CONCEPT FRAMEWORK
	**Primary Concept**: Faraday's Law of Electromagnetic Induction
	**Related Topics**: Motional EMF, Rotational Motion, Magnetic Force on a Moving Charge
	**Prerequisites**: Understanding of magnetic flux, induced EMF, and the concept of angular velocity.
## SOLUTION PATHWAY
**Given Information**:
	* A conducting disc rotating with angular velocity $\omega$
	* Uniform magnetic field $B$ perpendicular to the disc
	* Radius of the disc is $r$
	* A conducting ring is placed on the rim of the disc
**Method & Steps**:
	1.	Consider a small radial element of length $dr$ at a distance $x$ from the center of the disc.
		Reasoning: The induced emf is due to the motion of charges in the radial direction.
		Formula/Rule: The velocity of the element is $v = x\omega$
	2.	The induced emf across this element is $d\epsilon = vBdx$
		Working: Substituting $v = x\omega$, we get $d\epsilon = B\omega x dx$
		Key Point: The induced emf is due to the magnetic force on the moving charges.
	3.	Integrate this emf from the center of the disc ($x=0$) to the rim ($x=r$).
		Result: $\epsilon = \int_{0}^{r} B\omega x dx = B\omega [\frac{x^2}{2}]_{0}^{r} = \frac{1}{2}B\omega r^2$
		Verification: The derived expression matches the correct answer.
 ## OPTION ANALYSIS
[A] ✓: The induced emf is given by the integral of the motional emf across the radial length of the disc, which evaluates to  $\frac{1}{2}B\omega r^2$.
[B] :x: This is incorrect because the induced emf is proportional to $x^2$ and the integration leads to a factor of 1/2.
[C] :x: This is incorrect because the induced emf is proportional to $x^2$ and the integration leads to a factor of 1/2.
[D] :x: This is incorrect because the induced emf is proportional to $x^2$ and the integration leads to a factor of 1/2.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to integrate the induced emf over the length of the radius. [Always integrate the infinitesimal emf elements along the radial direction to obtain the total emf.]
		2.[Error 2]: Using the linear velocity at the rim ($r\omega$) directly for all points of the disc. [Remember that the linear velocity varies linearly from center to the rim.]
	**Quick Tips**:
		1.[Helpful shortcut]: Recall that the average velocity of the charges is half the velocity at the rim, which can quickly give an idea of the induced emf.
		2.[Memory aid]:  The induced emf in a rotating disc is proportional to the square of the radius and the angular velocity.

VENKATESWARARAO · Answer

The correct Option is "(1/2)Bωr²"

VENKATESWARARAO · Answer

Bωr²

VENKATESWARARAO · Answer

2Bωr²

VENKATESWARARAO · Answer

(1/4)Bωr²

NEET Physics Electromagnetic Induction Induced Electric Field, Eddy Currents MCQs

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