Induced Electric Fields MCQ Quiz in मल्याळम - Objective Question with Answer for Induced Electric Fields - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation:

The induced EMF produced in a moving rod is given by the formula:

EMF = B L v sin(theta)

where B is the magnetic field strength, L is the length of the rod, v is the velocity of the rod, and theta is the angle between the direction of motion of the rod and the magnetic field.

• If the magnetic field becomes parallel to the rails, then the angle between the direction of motion of the rod and the magnetic field is zero, i.e. theta = 0. In this case, the sine of theta becomes zero, and therefore the induced EMF becomes zero.
• So, if the magnetic field becomes parallel to the rails instead of being perpendicular, the induced EMF produced in the moving rod will be zero.

Calculation:

We can determine the induced current using Faraday’s Law of Electromagnetic Induction, which states:

\(\epsilon=-\frac{d\phi}{dt}\)

where:

\(\epsilon \text{ is the induced EMF (voltage),}\)

\(\phi \text{ is the magnetic flux, and }\)

\(\frac{d\phi}{dt} \text{ is the rate of change of flux. }\)

Given:

\(\phi= (2t^3+5t+7)mWb=(2t^3+5t+7)\times 10^{-3} Wb\)

Differentiate Magnetic Flux

\(\frac{d\phi}{dt}=\frac{d}{dt}(2t^3+5t+7)\)

\(\frac{d\phi}{dt}=6t^2+5\)

Step 2: Calculate at t=5s t = 5s" id="MathJax-Element-107-Frame" role="presentation" style="position: relative;" tabindex="0">t=5s$t = 5s$ $t = 5s$ .

\(\frac{d\phi}{dt}6(5)^2+5=6(25)+5=155mV=155\times 10^{-3}V\)

Ohm’s Law states:

\(I=\frac{\epsilon}{R}\)

Given R=10Ω R = 5 \Omega" id="MathJax-Element-108-Frame" role="presentation" style="position: relative;" tabindex="0">R=10Ω$R = 5 \Omega$ $R = 5 \Omega$ :

\(I=\frac{155\times 10^{-3}}{10}=15.5mA\)

Thus, option '2' is correct.

Concept:

According to Lenz’s law, the direction of the induced current is such that it opposes the change in magnetic flux causing it. When a conducting ring falls toward a long straight conductor carrying a current I:

• The magnetic field around the conductor decreases with increasing distance.
• As the ring approaches the conductor, the magnetic flux through the ring increases.
• The induced current in the ring will oppose this increase in flux by generating a magnetic field in the opposite direction.

Explanation:

Using the right-hand thumb rule for the current in the straight conductor:

• The magnetic field around the conductor forms concentric circles, pointing into the plane inside the ring.
• As the ring approaches the conductor, the flux through it increases. To counteract this, the induced current in the ring flows clockwise, producing a magnetic field out of the plane.

∴ The correct answer is: Induced electric current is clockwise.

The correct option is 3).

Calculation: 

At t = 10 sec complete loop is in magnetic field therefore no change in flux

e = \(\frac{\mathrm{d} \phi}{\mathrm{dt}}\) = 0

e = 0 for a complete loop 

∴ The value of induced emf in the loop at t = 10 s will be zero

Concept: 

For finding the magnitude of the magnetic moment of the electric field we should know how it varies with distance. 

We should check whether it is a monopole moment, dipole, or quadrupole moment.

First, we have to check its monopole moment which is known as sum of total charges i.e. m = ∑ Q_i 

where Q = charge placed 

If the monopole moment is zero then the system is independent on considering the frame of the origin or we can consider our origin at any point to make our calculation easier.

Electric dipole moment is = P = ∑ p_i = ∑ q.dx = charge × distance

If it is zero then we have to test its quadrupole moment for the electric field. 

Calculation:

Given:  two charges +q and -q placed at distance L.  

Electric monopole moment is = m = -q +q = 0

Here electric monopole moment becomes zero so we have to calculate its electric dipole moment.

We are free to consider origin anywhere because the monopole moment is zero.

Electric dipole moment is = P = charge × distance (with directions)

The electric field is towards for negative charge and away from the positive charge.

Dipole moment = P = q(L/2) (î) + (-q)(L/2)(-î) = qL (î)

so here, the dipole moment is non-zero. we know that for dipole moment electric field varies as E ∝ 1/R³ 

Hence Option 3) is correct.

Calculation:

Given that:

Velocity = 12 cm/sec = 0.12 m/sec

Length of Rod = 15 cm = 0.15 m

B = 0.5 T.

Resistance R = 180 mΩ =  0.18 Ω 

To calculate the power dissipated as heat, we need to first calculate the current flowing.

Since the induced emf is calculated as 

E = BVL

= 0.5 × 0.12 × 0.15

= 9 × 10^-3 Volts.

The current I would be equal to 

I = E/R

I = 9 × 10-3/(0.18)

= 50 × 10^-3 A

The power dissipation is equal to 

P = I²R

= (50 × 10-3 )² × 0.18

P = 4.5 × 10^-4 Watt 

The correct answer is option (4)

Explanation:

Power dissipation is given by the formula E²/R

Induced emf is given as 

E = BVL

Given that:

B = 0.5 T

V = 12 cm/s = 0.12 m/s

L = 15 cm = 0.15 m.

R = 180 mΩ = 0.18 Ω

Calculating E

E = 0.5 × 0.12 × 0.15 

= 9 × 10^-3 Volts

Power P = E2/R

= (9 × 10-3)²/0.18

= 4.5 × 10-4 W

The correct answer is option (3)