Q.A rectangular wire loop of sides $8\ \text{cm}$ and $2\ \text{cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3\ \text{T}$ directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1\ \text{cm s}^{-1}$ in a direction normal to the

Concept — Motional emf: $\varepsilon=B\,l\,v$, where $l$ is the length of the edge that cuts the field lines (the edge perpendicular to $\vec{v}$). The emf lasts only while the loop is leaving the field.

Data: $B=0.3\ \text{T}$, $v=1\ \text{cm s}^{-1}=0.01\ \text{m s}^{-1}$, sides $8\ \text{cm}=0.08\ \text{m}$ and $2\ \text{cm}=0.02\ \text{m}$.

(a) Velocity normal to the longer (8 cm) side. The cutting edge is the $8\ \text{cm}$ side, so $l=0.08\ \text{m}$:

The loop clears the field after travelling its $2\ \text{cm}$ width: $t_a=0.02/0.01=2\ \text{s}$. …

Principle. As a loop leaves a uniform field, only the edge still inside the field and perpendicular to the velocity acts as the seat of emf. That edge, of length $l$, gives a motional emf

and the emf persists only for as long as the loop is actually crossing the boundary — i.e. while it travels the dimension measured along its direction of motion.

Given: $B=0.3\ \text{T}$ (normal to the loop), $v=1\ \text{cm s}^{-1}=0.01\ \text{m s}^{-1}$, sides $8\ \text{cm}=0.08\ \text{m}$ and $2\ \text{cm}=0.02\ \text{m}$.

(a) Velocity normal to the longer (8 cm) side.

The velocity is perpendicular to the $8\ \text{cm}$ side, so that $8\ \text{cm}$ edge is the one cutting the field: $l=0.08\ \text{m}$.

The loop moves along its $2\ \text{cm}$ dimension to clear the field: …

Method: Motional EMF using $\varepsilon = B l v$

This is the standard method for calculating induced EMF when a conductor moves through a uniform magnetic field.

Concept (Why this works)

When a conductor of length $l$ moves with velocity $v$ perpendicular to a magnetic field $B$, the free electrons experience a magnetic force $F = q v B$. This force pushes charges along the conductor, creating an electric field and hence a potential difference — the motional EMF:

Here, $l$ is the length of the conductor that is actually cutting the field lines at that instant.

Steps for part (a): Velocity normal to the longer side (8 cm side)

1. Identify the cutting length The loop moves out of the field. The side that is cutting the field is the side perpendicular to velocity. Here, velocity is normal to the 8 cm side → the 2 cm side is the cutting length.

2. Apply motional EMF formula

3. Find duration of induced voltage The EMF lasts as long as the loop is partially inside the field. The loop must move a distance equal to its longer side (8 cm) to completely exit.

Steps for part (b): Velocity normal to the shorter side (2 cm side)

1. Identify the cutting length Velocity is normal to the 2 cm side → the 8 cm side is now cutting the field. $l = 8\ \text{cm} = 0.08\ \text{m}$ …

Common Mistakes & How to Avoid Them (Motional EMF)

Mistake 1: Using the wrong length for the moving conductor

The error: Students often take the side parallel to velocity as the effective length. For example, when velocity is normal to the longer side (8 cm), they incorrectly use 8 cm in the formula $\mathcal{E} = B l v$.

Why it's wrong: The motional EMF formula $\mathcal{E} = B l v$ uses $l$ = length of the conductor perpendicular to both $\vec{B}$ and $\vec{v}$. Only the side cutting the field lines matters.

How to avoid:

• Draw the loop and velocity vector.
• Identify which side is actually moving across the field lines.
• The effective length is the side perpendicular to velocity (and also perpendicular to $\vec{B}$).

Correct application:

• Case (a): Velocity normal to longer side → shorter side (2 cm) cuts the field → $l = 2\ \text{cm}$
• Case (b): Velocity normal to shorter side → longer side (8 cm) cuts the field → $l = 8\ \text{cm}$

Mistake 2: Forgetting to convert units

The error: Using cm directly without converting to metres. This gives answers off by a factor of 100.

Why it's wrong: The SI unit of magnetic field is Tesla ($T = \frac{N}{A \cdot m}$). Length must be in metres, velocity in m/s.

How to avoid:

• Always convert: $1\ \text{cm} = 10^{-2}\ \text{m}$
• Write conversion step explicitly:
  • $8\ \text{cm} = 0.08\ \text{m}$
  • $2\ \text{cm} = 0.02\ \text{m}$
  • $1\ \text{cm/s} = 0.01\ \text{m/s}$

Correct calculation:

• Case (a): $\mathcal{E} = (0.3)(0.02)(0.01) = 6 \times 10^{-5}\ \text{V}$
• Case (b): $\mathcal{E} = (0.3)(0.08)(0.01) = 2.4 \times 10^{-4}\ \text{V}$

Mistake 3: Confusing the duration of induced EMF

The error: Students think the EMF lasts as long as the loop is partially inside the field, but they calculate the time incorrectly — often using the wrong side length.

Why it's wrong: The EMF exists only while one side is inside the field and the opposite side is outside — i.e., while the loop is entering or exiting. The time depends on the length of the side parallel to velocity.

How to avoid:

• The EMF lasts until the trailing side completely exits the field.
• Time = $\frac{\text{length of side parallel to velocity}}{\text{speed}}$

Correct application:

• Case (a): Velocity normal to longer side → longer side (8 cm) is parallel to velocity → $t = \frac{0.08}{0.01} = 8\ \text{s}$
• Case (b): Velocity normal to shorter side → shorter side (2 cm) is parallel to velocity → $t = \frac{0.02}{0.01} = 2\ \text{s}$

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