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

Question

A conducting rod of length $l$ rotates with a constant angular velocity $\omega$ about one end in a uniform magnetic field $B$ perpendicular to the plane of rotation.  A capacitor $C$ is connected in series with the rod and a resistance $R$.  If the steady state charge on the capacitor is $Q$, which of the following expressions is correct?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Motional EMF and Capacitor Charging
	**Related Topics**: Faraday's Law of Induction, Capacitance, Electric Potential
	**Prerequisites**: Understanding of motional EMF, capacitance, and series circuits
## SOLUTION PATHWAY
**Given Information**:
	* Length of the rod: $l$
	* Angular velocity: $\omega$
	* Magnetic field: $B$
	* Capacitance: $C$
	* Steady-state charge on capacitor: $Q$
	* The magnetic field is perpendicular to the plane of rotation.
**Method & Steps**:
	1. Calculate the motional EMF induced in the rotating rod.
		Reasoning: The rotating rod acts as a source of EMF due to the motion of charges in the magnetic field.
		Formula/Rule: Motional EMF formula: $E = \frac{1}{2}B\omega l^2$
	2. Determine the voltage across the capacitor in steady state.
		Working: In steady state, the capacitor is fully charged, and no current flows. Thus, the voltage across the capacitor equals the induced EMF in the rod.
		Key Point: In steady state, current in the circuit is zero.
	3. Relate the charge on the capacitor to the voltage across it.
		Result: The charge on the capacitor is given by $Q = CV$, where $V$ is the voltage across the capacitor. Substituting the EMF for V, we get $Q = C( \frac{1}{2}B\omega l^2) = \frac{1}{2}CB\omega l^2$
		Verification: The units of the result match the unit of charge.
 ## OPTION ANALYSIS
[A] ✓: The correct expression for the charge on the capacitor is $Q = \frac{1}{2}CB\omega l^2$, which matches option A.
[B] :x: Option B is missing the factor of 1/2, hence it is incorrect.
[C] :x: Option C has a factor of 2 instead of 1/2, making it incorrect.
[D] :x: Option D has a factor of 1/4 instead of 1/2, making it incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting the 1/2 factor in the motional EMF formula for a rotating rod. [How to avoid]: Always remember that the motional EMF for a rotating rod is given by $E = \frac{1}{2}B\omega l^2$ and not $B\omega l^2$.
		2.[Error 2]: Confusing the voltage across the capacitor with the total EMF in the circuit. [How to avoid]: In steady state, when the capacitor is fully charged, the voltage across the capacitor equals the induced EMF. 
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the general motional EMF formula, $EMF = vBL$, and then use integration to find the EMF in a rotating rod, resulting in the factor of 1/2.
		2.[Memory aid]: The rotating rod's EMF is always less than that of a rod moving at the same speed at its end, it is half the value.

SUBHANI · Answer

The correct Option is "Q = 1/2 CBωl^2"

SUBHANI · Answer

Q = CBωl^2

SUBHANI · Answer

Q = 2CBωl^2

SUBHANI · Answer

Q = 1/4 CBωl^2

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