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

Question

A uniform circular metal disc of radius ‘R’ is rotating about a vertical axis passing through its center and perpendicular to its plane with constant frequency f. If ${B_H}$ and ${B_V}$ are horizontal and vertical components of the earth’s magnetic field respectively, then the induced emf between its centre and the rim is

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Motional EMF due to a rotating conductor in a magnetic field
	**Related Topics**: Faraday's Law of Induction, Magnetic Flux
	**Prerequisites**: Understanding of magnetic fields, circular motion, and induced EMF
## SOLUTION PATHWAY
**Given Information**:
	* Radius of the disc: R
	* Frequency of rotation: f
	* Horizontal component of magnetic field: B_H
	* Vertical component of magnetic field: B_V
**Method & Steps**:
	1. Identify the relevant magnetic field component: Only the vertical component ($B_V$) of the magnetic field will contribute to the induced EMF in the rotating disc. The horizontal component will produce a force parallel to the disc, not inducing an EMF between center and rim
		Reasoning: The motional EMF is induced when a conductor cuts magnetic field lines. In this case, the disc rotates in the plane and only the magnetic field perpendicular to the plane ($B_V$) will lead to an induced EMF.
	2. Calculate the motional EMF using the formula: The induced EMF (ε) for a rotating rod of length R in a magnetic field B with angular velocity ω, is given by ε = 1/2 * B * ω * R^2. Since angular velocity (ω) = 2πf, the formula can be modified to ε = 1/2 * B * 2πf * R^2.
		Formula/Rule:  ε = 1/2 * B * ω * R^2 and ω = 2πf 
	3. Substitute the values and simplify: Substituting B_V for B and 2πf for ω, we get  ε = 1/2 * B_V * 2πf * R^2 = πB_VfR^2
		Result: ε = πB_VfR^2
		Verification: The derived formula matches with the correct option (A). 
 ## OPTION ANALYSIS
[A] ✓: Correct, as it matches the derived result πB_VfR^2.
[B] :x: Incorrect, as the horizontal component of magnetic field (B_H) does not contribute to the induced EMF in the disc.
[C] :x: Incorrect, the factor of 2 is wrong.
[D] :x: Incorrect, as there is an induced EMF due to the rotating disc in the vertical magnetic field.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Using B_H instead of B_V: Confusing the horizontal and vertical components of the magnetic field. Always consider the component perpendicular to the motion of the conductor.
		2. Forgetting the factor of 1/2 in the EMF formula: The factor 1/2 is crucial for the rotating disc case; it does not exist for linear motion of rod
	**Quick Tips**:
		1. Remember that the induced EMF in a rotating conductor is due to the component of the magnetic field perpendicular to the plane of rotation.
		2. Visualize the direction of the force on charges due to the magnetic field to confirm the direction of the induced EMF.

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\pi {B_v}f{R^2}$"

Mr. ANIL KUMAR.G · Answer

$\pi {B_H}f{R^2}$

Mr. ANIL KUMAR.G · Answer

$2\pi {B_v}f{R^2}$

Mr. ANIL KUMAR.G · Answer

zero

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