Motional Electromotive Force MCQ Quiz in मल्याळम - Objective Question with Answer for Motional Electromotive Force - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

• ​EMF (Electromotive Force) ε is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell.
• It is the force which causes to flow the free electrons in any closed circuit due to difference in electrical pressure or potential.
• It is represented by ‘E’. Its unit is Volt.
• It is nothing but the potential difference.

Important Points

Side length \( \Rightarrow s=5 \, \text{cm}=5 \times 10^{-2} \, \text{m} \)

Magnetic field \( \Rightarrow B=0.6 \, \text{T} \)

Refer image,

Here, three cases will exist

1) When block is outside the magnetic field

Initially emf \( \Rightarrow e=0 \)

2) When it starts entering the magnetic field

\( \therefore \text{emf} \Rightarrow e=BLV= 0.6 \times 5 \times 10^{-2} \times 1 \times 10^{-2}=3 \times 10^{-4} \)

\( E=-3 \times 10^{-4} \)

Above emf will have a negative sign as flux will oppose in the opposite direction of the magnetic field.

3) When block is in the magnetic field.

After \( t=5 \, \text{s} \) it has covered \( 5 \, \text{cm} \) (length)

\( \therefore 15 \, \text{cm} \) is remaining. .................[Magnetic field length = 2 cm]

\( \therefore \) From the above 3 cases,

Initially, emf will be zero then emf will be induced with

emf of \( -3 \times 10^{-4} \) and then covering \( 20 \, \text{cm} \)

it will induce emf \( =+3 \times 10^{-4} \)

\( \therefore \) Option c is correct.

Given:

Square side length \(a = \frac{1}{\sqrt{2}} \, \text{m current } I = 5 \, \text{A} .\)

To find the magnetic field ( B ) at the center O of the square.

Concept:

• The magnetic field at the center of a square due to one side carrying current is given by:
• \(B_{side} = \frac{\mu_0 I}{4 \pi R} \left(\sin\theta_1 + \sin\theta_2\right)\) , where R is the perpendicular distance from the center of the square to the side, and \(\theta_1\) , \(\theta_2\) are the angles subtended by the side at the center.
• For a square, each side contributes equally, and the total field is four times the field due to one side.

Calculation:

The distance from the center of the square to the midpoint of a side:

\(⇒ R = \frac{a}{2} = \frac{1}{2\sqrt{2}} \, \text{m} .\)

For each side, \(\theta_1 = \theta_2 = 45^\circ , so \sin\theta_1 + \sin\theta_2 = 2 \sin 45^\circ = \sqrt{2} .\)

Magnetic field due to one side:

\(⇒ B_{side} = \frac{\mu_0 I}{4 \pi R} \cdot \sqrt{2} .\)

Substituting \(\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} :\)

\(⇒ B_{side} = \frac{4\pi \times 10^{-7} \cdot 5}{4\pi \cdot \frac{1}{2\sqrt{2}}} \cdot \sqrt{2} . \\ ⇒ B_{side} = 2\times 10^{-6} \, \text{T} = 2 \, \mu\text{T} .\)

Total magnetic field at the center (from 4 sides):

\(⇒ B_{total} = 4 \cdot B_{side} = 4 \cdot 2\, \mu\text{T} = 8 \, \mu\text{T} .\)

∴ The magnetic field at the center is \(4 \, \mu\text{T}\) .

Explanation:

The problem involves a metal rod moving through a magnetic field and inducing an electromotive force (EMF). The rod's length, velocity, magnetic field strength, and the angle at which it moves through the magnetic field are given. We need to find the magnitude of the induced EMF.

First, let's summarize the given data:

• Length of the rod (l): 1.5 meters
• Velocity of the rod (v): 10 meters per second
• Magnetic field strength (B): 2 Tesla
• Angle between the rod's motion and the magnetic field (θ): 60 degrees

To find the magnitude of the induced EMF, we can use the formula for the motional EMF induced in a conductor moving through a magnetic field:

EMF = Blv sin(θ)

Where:

• EMF is the induced electromotive force
• B is the magnetic field strength
• l is the length of the conductor
• v is the velocity of the conductor
• θ is the angle between the direction of motion and the magnetic field

Let's plug in the given values into the formula:

EMF = 2 T x 1.5 m x 10 m/s x sin(60 degrees)

We know that sin(60 degrees) = √3/2.

Therefore:

EMF = 2 x 1.5 x 10 x (√3/2)

EMF = 30 x (√3/2)

EMF = 15√3 volts

Let's calculate the numerical value of 15√3:

√3 ≈ 1.732

EMF ≈ 15 x 1.732

EMF ≈ 25.98 volts

Rounding to two decimal places, we get approximately 26 volts.

Therefore, the magnitude of the induced EMF is approximately 26 volts, which matches option 4.

\(de = B(\omega x) \cdot dx\)

e = B\omega \int_{2L}^{3L} x \, dx

= \dfrac{5B\omega L^{2}}{2}

Side length \( \Rightarrow s=5 \, \text{cm}=5 \times 10^{-2} \, \text{m} \)

Magnetic field \( \Rightarrow B=0.6 \, \text{T} \)

Refer image,

Here, three cases will exist

1) When block is outside the magnetic field

Initially emf \( \Rightarrow e=0 \)

2) When it starts entering the magnetic field

\( \therefore \text{emf} \Rightarrow e=BLV= 0.6 \times 5 \times 10^{-2} \times 1 \times 10^{-2}=3 \times 10^{-4} \)

\( E=-3 \times 10^{-4} \)

Above emf will have a negative sign as flux will oppose in the opposite direction of the magnetic field.

3) When block is in the magnetic field.

After \( t=5 \, \text{s} \) it has covered \( 5 \, \text{cm} \) (length)

\( \therefore 15 \, \text{cm} \) is remaining. .................[Magnetic field length = 2 cm]

\( \therefore \) From the above 3 cases,

Initially, emf will be zero then emf will be induced with

emf of \( -3 \times 10^{-4} \) and then covering \( 20 \, \text{cm} \)

it will induce emf \( =+3 \times 10^{-4} \)

\( \therefore \) Option c is correct.

Concept:

Induced EMF in a Moving Loop in a Magnetic Field:

• According to Faraday's law of induction, the induced EMF (e) in the loop is given by e = -dΦ/dt, where Φ is the magnetic flux through the loop.
• The magnetic flux is given by Φ = B × A, where B is the magnetic field and A is the area of the loop.
• As the loop moves into the magnetic field, the area A increases, resulting in a changing magnetic flux, which induces an EMF.
• The speed of the loop is constant (2 cm/s), and the magnetic field strength is 0.5 T.
• The induced EMF is directly proportional to the speed and the area of the loop that enters the magnetic field.

Calculation:

Given:

Magnetic field, B = 0.5 T

Speed of the loop, v = 2 × 10^-2 m/s

Side of the square loop, a = 2 cm = 2 × 10^-2 m

The induced EMF when the front edge enters the field:

e = B × l × v

e = 0.5 × 2 × 10^-2 × 2 × 10^-2

e = 2 × 10^-4 V

The induced EMF is constant as long as the loop moves with a constant speed and the magnetic field is uniform. The graph showing the induced EMF is a step function, starting from 0 and increasing linearly, as the loop enters the field.

∴ The correct answer is: Option 3 (The graph showing the induced EMF as a constant value after a certain time).

Concept Used:

 When a conductor moves perpendicular to a magnetic field, an emf is induced in it.
The emf (ε) induced in a rod of length L moving with velocity v perpendicular to a magnetic field B is given by: ε = B × L × v

Calculation:

Velocity of the rod after 2 seconds:
     v = g × t
     v = 10 m/s² × 2 s
     v = 20 m/s
⇒ Induced emf:
     ε = B × L × v
     ε = 3 × 10^-5 T × 1 m × 20 m/s
⇒ ε = 6 × 10^-4 V

∴ The emf induced in the rod at t = 2 s is 6 × 10^-4 V.

Concept:

Motional emf is the emf induced in a conducting rod when it moves in a region of the magnetic field.

Mathematically, 

Motional EMF ⇒ ϵ = BVL

where, B = magnetic induction, V = velocity and L = length

Calculation:

Motional EMF

ϵ = BVL

= (v×B).L

ϵ = BVL = IR

I  =  \({BVL\over R}\)

F = ILB

F_B = I (L×B)

F = F_B

ILB = F

F = \({BVL\over R}\)LB

∴ F =  B2L2 v/R

The correct answer is option (3).

Calculation: 

The Earth's magnetic field is not fully perpendicular to the plane of the rotating fan. The vertical component of the magnetic field that is perpendicular to the plane of rotation is:

⇒ \(B_v=B \sin 30=\frac{1}{4} × 10^{-4}\)

⇒ \(\omega=2 π × f=\frac{2 π}{60} × 1200 \mathrm{rad} / \mathrm{s}\)

⇒ \(\varepsilon=\frac{1}{2} B_V \omega \ell^2\)

⇒ 32π × 10^-5 V

∴  The value of N is 32.