A region in the form of an equilateral triangle (in x-y plane) of height L has a uniform magnetic field \vecB pointing in the +z -direction. A conducting loop PQR , in the form of an equilateral triangle of the same height L , is placed in the x-y plane with its vertex P at x=0 in the orientation shown in the figure. At t=0 , the loop starts entering the region of the magnetic field with a uniform velocity \vecv along the +x -direction. The plane of the loop and its orientation remain unchanged throughout its motion. Which of the following graph best depicts the variation of the induced emf (E) in the loop as a function of the distance (x) starting from x=0 ?

Electromagnetic Induction PYQ — JEE Advanced 2024 · Shift 2 · Q35 with Solution

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Correct answer: A

Key idea: induced emf depends on rate of change of overlapped area

Since the magnetic field is uniform and perpendicular to the loop,

E = \left|\frac{d\Phi}{dt}\right| = B\left|\frac{dA}{dt}\right|.

As the loop moves with constant speed $v$ along $+x$ ,

\frac{dA}{dt} = \frac{dA}{dx}\frac{dx}{dt} = v\frac{dA}{dx}.

Hence,

E = Bv\left|\frac{dA}{dx}\right|.

So we only need to find how the common area between the two equilateral triangles changes with horizontal displacement $x$ .

Geometry of the triangles

Both the magnetic-field region and the loop are equilateral triangles of the same height $L$ .

For an equilateral triangle of height $L$ , if measured from its vertex along the horizontal direction, the vertical width increases linearly from $0$ at the vertex to maximum at the base.

Thus, during entry, the overlapped part is itself a similar triangular region whose dimensions grow linearly with $x$ .

So the overlap area grows as

A(x) \propto x^2

for the first part of motion. Therefore,

\frac{dA}{dx} \propto x

which means

E \propto x.

So the induced emf increases linearly from zero.

What happens after further motion?

As the loop penetrates more, the overlap no longer remains a growing triangle. Because the two triangles are congruent and identically oriented, after a certain point the rate at which area is added starts decreasing symmetrically.

Thus:

emf starts from $0$ ,
increases linearly,
reaches a maximum at the symmetric configuration,
then decreases linearly back to $0$ .

So the $E$ vs $x$ graph is a symmetric triangular graph.

Why not constant / curved / discontinuous?

Not constant: because overlap width changes with position.
Not curved: because $A \propto x^2$ in the entry part, so $dA/dx \propto x$ , i.e. linear.
No sudden jumps: geometry changes smoothly.

Correct option

Hence the graph must be the one where emf:

starts at zero,
rises linearly,
then falls linearly to zero.

Therefore, the correct option is

\boxed{A}

Comparison with stored answer

Stored correct answer: $A$

My derived answer: $A$

So they agree.

More from Electromagnetic Induction

View written solutionFree

Correct answer: A

Key idea: induced emf depends on rate of change of overlapped area

Since the magnetic field is uniform and perpendicular to the loop,

E = \left|\frac{d\Phi}{dt}\right| = B\left|\frac{dA}{dt}\right|.

As the loop moves with constant speed $v$ along $+x$ ,

\frac{dA}{dt} = \frac{dA}{dx}\frac{dx}{dt} = v\frac{dA}{dx}.

Hence,

E = Bv\left|\frac{dA}{dx}\right|.

So we only need to find how the common area between the two equilateral triangles changes with horizontal displacement $x$ .

Geometry of the triangles

Both the magnetic-field region and the loop are equilateral triangles of the same height $L$ .

For an equilateral triangle of height $L$ , if measured from its vertex along the horizontal direction, the vertical width increases linearly from $0$ at the vertex to maximum at the base.

Thus, during entry, the overlapped part is itself a similar triangular region whose dimensions grow linearly with $x$ .

So the overlap area grows as

A(x) \propto x^2

for the first part of motion. Therefore,

\frac{dA}{dx} \propto x

which means

E \propto x.

So the induced emf increases linearly from zero.

What happens after further motion?

Thus:

emf starts from $0$ ,
increases linearly,
reaches a maximum at the symmetric configuration,
then decreases linearly back to $0$ .

So the $E$ vs $x$ graph is a symmetric triangular graph.

Why not constant / curved / discontinuous?

Not constant: because overlap width changes with position.
Not curved: because $A \propto x^2$ in the entry part, so $dA/dx \propto x$ , i.e. linear.
No sudden jumps: geometry changes smoothly.

Correct option

Hence the graph must be the one where emf:

starts at zero,
rises linearly,
then falls linearly to zero.

Therefore, the correct option is

\boxed{A}

Comparison with stored answer

Stored correct answer: $A$

My derived answer: $A$

So they agree.

Electromagnetic Induction question

Electromagnetic Induction question

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Electromagnetic Induction question

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