Test: MCQs (One or More Correct Option): Electromagnetic Induction & Alternating Current | JEE Advanced - JEE MCQ

KEY CONCEPT : The magnetic field due to a current flowing in a wire of finite length is given by

Applying the above formula for AB for finding the field at O, we get

acting perpendicular to the plane of paper upwards

∴ The total magnetic field due to current flowing through ABCD is

The total flux passing through the square EFGH

[∵ ℓ > L and therefore, B can be assumed constant for ℓ²]

The flux through small square loop is directly proportional to the current passing through big square loop.

∴ φ₂ ∝ I₁ ⇒ φ = MI₁ where M = Mutual Conductance

NOTE : Since the rate of change of magnetic flux is zero, hence there will be  no net induced emf and hence no current flowing in the loop.

Rate of change of current 

Induced emf 

Power  P = V₂ i₂ = 2 × 10 ^–3 × m × i₂

Since Power is equal

∴ 8 × 10 ^{– 3} × mi₁ = 2 ×10 ^{– 3} mi₂

      ...(i)

KEY CONCEPT : The magnetic field due to a current flowing in a wire of finite length is given by

Applying the above formula for AB for finding the field at O, we get

acting perpendicular to the plane of paper upwards

∴  The total magnetic field due to current flowing through ABCD is

The total flux passing through the square EFGH

[∵ ℓ  > L and therefore, B can be assumed constant for ℓ²]

The flux through small square loop is directly proportional to the current passing through big square loop.
∴ φ₂ ∝ I₁ ⇒ φ = MI₁ where M = Mutual Conductance

∴ L = R.t or henry  = ohm-second.

A motional emf, e = Blv is induced in the rod. Or we can say a potential difference is induced between the two ends of the rod AB, with A at higher potential and B at lower potential.
Due to this potential difference, there is an electric field in the rod.

The capacitance in case B is four times the capacitance in case A

∴ Impedance in case B is less then that of case A (Z_B < Z_A)

 option (a) is correct.

[∵ If V is the applied potential difference access series R-C circuit then V 

∴ (c) is the correct option.

Impedance across AB

Impedance across CD is

∴The current I from the circuit is

Emf will be induced in the circular wire loop when flux through it changes with time.

when the current is constant, the flux changing through it will be zero.

When the current is decreasing at a steady rate then the change in the flux (decreasing inwards) on the right half of the wire is equal to the change in flux (decreasing outwards) on the left half of the wire such that Δφ through the circular loop is zero.

I = cos 500 t

 the charge will be maximum at 

From the graph it is clear that just before  the
current is in anticlockwise direction.

∴ (b) is incorrect

At  the charge on the upper plate of capacitor
is

Now applying KVL (when A is just connected to D)

∴ (c) is the correct option.
The maximum charge on C is Q = CV = 20 × 10^–6 × 50 =10^–3C
Therefore, the total charge flown = 2 × 10^–3 C

∴ (d) is the correct option.

The flux passing through the triangular wire if i current flows through the inifinitely long conducting wire

Induced emf  in the wire 

As the current in the triangular wire is decreasing the induced current in AB is in the same direction as the  current in the hypotenuse of the triangular wire. Therefore force will be repulsive.

 [Counter clockwise directionwhile entering, Zero when completely inside and clockwise while exiting]

 [Toward left while entering and exiting and zero when completely inside]

[V decreases from x = 0 to x = L, remains constant for x = L to x= 3L again decreases from x=3L to x=4L] From (i) and (iii)

[i decreases from x = 0 to x = L i becomes zero from x = L to x = 3L i changes direction and decreases from x = 3L to x = 4L]

These characteristics are shown in graph (a) and (b) only.