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

Question

A conducting square loop of side $l$ and resistance $R$ is moving with a constant velocity $v$ perpendicular to one of its sides and entering a uniform magnetic field $B$ perpendicular to the plane of the loop. What is the magnitude of the induced current in the loop as a function of time $t$, where $t=0$ is the instant when the loop starts entering the field?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Faraday's Law of Induction and Ohm's Law
 **Related Topics**: Electromagnetic induction, magnetic flux, induced EMF
 **Prerequisites**: Understanding of magnetic flux, Faraday's law, and Ohm's law.
## SOLUTION PATHWAY
**Given Information**:
  * Side of the square loop: $l$
  * Resistance of the loop: $R$
  * Velocity of the loop: $v$
  * Magnetic field strength: $B$
  * Time: $t = 0$ when the loop starts entering the field.
  * The loop is moving perpendicular to one of its sides and entering the field perpendicular to its plane.
**Method & Steps**:
 1. **Determine the magnetic flux:** The magnetic flux ($\Phi$) through the loop changes as it enters the magnetic field.  At time $t$, the area of the loop within the field is $A(t) = l(vt)$.  Therefore, the magnetic flux is given by $\Phi(t) = B 	imes A(t) = Bl(vt) = Blvt$ for $0 < t < \frac{l}{v}$.
  Reasoning: The flux is the product of the magnetic field and the area of the loop inside the field.
  Formula/Rule: $\Phi = BA$
 2. **Calculate the induced EMF:** According to Faraday's law of induction, the induced EMF ($\epsilon$) is the negative rate of change of magnetic flux with respect to time:
  $\epsilon(t) = -\frac{d\Phi(t)}{dt} = -\frac{d}{dt}(Blvt) = -Blv$ for $0 < t < \frac{l}{v}$. The negative sign indicates the direction of the induced current (Lenz's Law), but we're only interested in the magnitude.
  Working: Differentiating the flux with respect to time.
  Key Point: The induced EMF is constant while the loop is entering the field.
 3. **Calculate the induced current:** Using Ohm's law, the induced current ($I$) is the induced EMF divided by the resistance:
  $I(t) = \frac{\epsilon(t)}{R} = \frac{Blv}{R}$ for $0 < t < \frac{l}{v}$. Once the loop is completely inside the field ($t \ge \frac{l}{v}$), the flux becomes constant, and the induced current becomes 0.
  Result: $I = \frac{Blv}{R}$ for $0 < t < \frac{l}{v}$, and $I = 0$ otherwise.
  Verification: The units are consistent (Amperes). The result makes physical sense; a constant current flows while the loop is partially in the field.
 ## OPTION ANALYSIS
A] ✓: This option correctly describes the induced current as a function of time.  The current is constant while the loop is entering the field and zero otherwise.
B] :x: This option is incorrect. The current doesn't depend inversely on time.
C] :x: This option is incorrect. The current is not constant for all time; it becomes zero once the loop is fully inside the field.
D] :x: This option is incorrect.  A current is induced while the loop is entering the field.
 ## LEARNING AIDS
 **Common Mistakes**:
  1. [Forgetting Lenz's Law]: Remember to consider the direction of the induced current, although only the magnitude was asked here.
  2. [Incorrectly calculating the changing area]: Make sure to correctly express the area of the loop within the magnetic field as a function of time.
 **Quick Tips**:
  1. [Visualize the problem]: Draw a diagram to help understand how the area of the loop inside the field changes with time.
  2. [Remember Faraday's and Ohm's law]: These are fundamental laws in electromagnetism and should be memorized.

VENKATESWARARAO · Answer

The correct Option is "$I = \frac{Blv}{R}$ for $0 < t < \frac{l}{v}$, and $I = 0$ otherwise"

VENKATESWARARAO · Answer

$I = \frac{B l^2 v}{Rt}$

VENKATESWARARAO · Answer

$I = \frac{Blv}{R}$ for all $t$

VENKATESWARARAO · Answer

$I = 0$

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