Induced Electric Fields MCQ Quiz in বাংলা - Objective Question with Answer for Induced Electric Fields - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 30, 2025
Latest Induced Electric Fields MCQ Objective Questions
Top Induced Electric Fields MCQ Objective Questions
Induced Electric Fields Question 1:
In the figure, a conducting ring of certain resistance is falling towards a current carrying straight long conductor. The ring and conductor are in the same plane. Then the
Answer (Detailed Solution Below)
Induced Electric Fields Question 1 Detailed Solution
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).
Induced Electric Fields Question 2:
A square loop of side 15 cm being moved towards right at a constant speed of 2 cm/s as shown in figure. The front edge enters the 50 cm wide magnetic field at t = 0. The value of induced emf in the loop at t = 10 s will be :
Answer (Detailed Solution Below)
Induced Electric Fields Question 2 Detailed Solution
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
Induced Electric Fields Question 3:
Two point charges -q and +q are placed at a distance of L, as shown in the figure.
The magnitude of electric field intensity at a distance R(R >> L) varies as:
Answer (Detailed Solution Below)
Induced Electric Fields Question 3 Detailed Solution
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 = ∑ Qi
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 = ∑ pi = ∑ 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.
Induced Electric Fields Question 4:
Comprehension:
Figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 180.0 mΩ. Assume the field to be uniform.
The power dissipated as heat in the closed circuit is:
Answer (Detailed Solution Below)
Induced Electric Fields Question 4 Detailed Solution
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)
Induced Electric Fields Question 5:
Comprehension:
Figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 180.0 mΩ. Assume the field to be uniform.
The power required (by an external agent) to keep the rod moving at the same speed (= 12 cms-1) when K is closed will:
Answer (Detailed Solution Below)
Induced Electric Fields Question 5 Detailed Solution
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)