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(a) a direct current flows in the ammeter A. 

M = 2.5H

dl/dt = lA/S

E = -μdl/dt

=> E = 2.5 x 1 = 2.5V

Correct option is (A) \(\cfrac12\)Bωl2

From the relation,
φ = LI,
it follows that the self inductance of a conductor is equal to the slope of its graph between φ and I. Therefore, the self inductance of the conductor A has the larger value.

The induced emf produced across the two ends of the conductor
e = \(\frac{1}{2}\) Bl² W = \(\frac{1}{2}\) × 0.2 × 10^-4 × l² × 5
= 5 ×10^-5V = 50 µV
Answer is (B) 50μV

The answer is (B) Relative position and orientation of the two coils

For the magnet, eddy currents are produced in the metallic pipe. These currents will oppose the motion of the magnet. Therefore magnet’s downward acceleration will be less than the acceleration due to gravity g. On the other hand, an unmagnetised iron bar will not produce eddy currents and will fall with an acceleration g. Thus the magnet will take more time.

Consider a long air-cored solenoid of length Z, diameter d and N turns of wire. We assume that the length of the solenoid is much greater than its diameter so that the magnetic field inside the solenoid may considered to be uniform, that is, end effects in the solenoid can be ignored. With a steady current I in the solenoid, the magnetic field within the solenoid is

B = µ₀nI ………….. (1)

where n = N/l is the number of turns per unit length. So the magnetic flux through one turn is 

Φ_m = BA = µ₀nIA ……….. (2)

Hence, the self inductance of the solenoid,

L = \(\cfrac{NΦ_m}l\) = (nl)µ₀nA = µ₀n² lA = µ₀n² V …………..(3)

= µ₀n² l \(\cfrac{πd^2}4\)…………. (4)

where V = lA is the interior volume of the solenoid. Equation (3) or (4) gives the required expression.

[Note: It is evident thatthe self inductance of a long solenoid depends only on its physical properties – such as the number of turns of wire per unit length and the volume, and not on the magnetic field or the current. This is true for inductors in general.] .

A dc ammeter cannot read ac because, the average value of ac is zero over a complete cycle.

Capacitive reactance, X_C = \(\frac{1}{ωC}\) = \(\frac{1}{2πƒc}\)

where, ƒ = 0, X_C = ∞

where, ƒ is the frequency of the ac supply. In a dc circuit ƒ = 0. Hence the capacitive reactance has infinite value for dc and a finite value for ac. In other words, a capacitor serves as a block for dc and offers an easy path to ac.

In rectangular coil the induced emf will remain constant because in this the case rate of change of area in the magnetic field region remains constant, while in circular coil the rate of change of area in the magnetic field region is not constant.

Correct option is  (D) 12π² V.

dϕ/dt = 8 x 10^-4weber

n = 200, l = 4A, E = -nL(dl/dt)

or, -dϕ/dt = -Ldl/dt

or, L = ndϕ/dt = 200 x 8 x10^-4 = 2 x 10^-2H.

(B) the plane of the coil is parallel to the magnetic field

An emf induced in a conductor or circuit moving in a magnetic field is called motional emf

Data: l = 88 cm, 

B = 2.5 Wb/m², 

B = 3 Wb/m², 

∆t = 0.5 s

For the circular loop, l = 2πr

∴ r = \(\cfrac{l}{2\pi}\) = \(\cfrac{88}{2\times(22/7)}\) = 14 cm = 0.14 m

Area of the circular loop, A_i = πr²

= \(\cfrac{22}7\)(0.14)² = 0.0616 m²

Initial magnetic flux, Φ_i = A_iB_i

= 0.0616 × 2.5 = 0.154 Wb 

For the square loop, length of each side

= \(\cfrac{88}4\) cm = 22 cm = 0.22 m⁴

Area of the square loop, A_f = (0.22)²

= 0.0484 m²

∴ Final magnetic flux, Φ_f = A_fB_f

= 0.0484 × 3 = 0.1452 Wb

Induced emf, e = – \(\cfrac{Φ_f−Φ_1}{Δt}\) = \(\cfrac{Φ_1−Φ_f}{Δt}\)

∴ e = \(\cfrac{0.154 - 0.1452}{0.5}\) =8.8 x 10^-3 x 2

= 1.76 × 10^-2 V

Correct option is  (B) 5ω mA

Correct option is (C) 1 N

a. The number of turns per unit length in each coil, 

b. Area of the coils, 

c. Length of the coils, 

d. permeability of medium inside the coils, 

e. separation between the coils, 

f. the relative orientation of the coils.  

EMF also increases (directly proportional to the number of turns). 

Decreases. The eddy currents induced in the conductor damp the motion of the conductor.

Increases (in all of the three cases) 

(b) 0.1 T

ε = Blv 

⇒ B =\(\frac{ε}{lv}\) = \(\frac{1}{1 \times 10}\) = 0.02 T

When a bulk conductor is placed in a varying magnetic field, circulating currents are induced in it. These currents are called eddy currents. 

No. Because glass is an insulator. 

No, induced emf (motional emf) in a conductor if it moves in a plane parallel to a magnetic field.

Unit of self-inductance is Henry.

No. (If it is made to fall in E – W direction, there is an emf across its ends) 

Fleming’s right hand rule is otherwise called generator rule.

The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the Electromagnetic Induction and Alternating Current thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.

No, because there is no change in magnetic flux.

Due to the opposing nature of growth or decay of current in the coil.

In Wheatstone bridge, initially a cell key and later on galvanometer key is pressed. If the galvanometer’s key is pressed first then the induced current flow in the coil reduce the main current. So the error is occured in the measurement.

Due to the effect of eddy currents in conductor; non metal coil will reach on the surface of Earth before the metal coin.

Coefficient of self induction is given by the relation:
L = \(\frac{\mu_{0} N^{2} A}{l}\)

The effect of eddy currents is reduced by using soft iron core.

Coefficient of self induction of coil is given by the relation : L = \(\frac{N^{2} \mu_{0} A}{l}\)
∴ L ∝ A so the self induction is 2 times.

A will become positive w.r.t. B as current induced is in a clockwise direction.

induced current is in anti-clockwise direction when seen from left hand side and its direction is in clockwise when seen from right hand side. Thus, direction of induced current is in clockwise sense.
This implies that plate A of the capacitor is at the higher potential than plate B, i.e. B will be negative plate, while A will be positive plate.

Eddy currents:

When a thick metallic piece is placed in a time varying magnetic field, the magnetic flux linked with the plate changes, the induced currents are set up in the conductor; these currents are called eddy currents. These currents are sometimes so strong, that the metallic plate becomes red hot.

Due to heavy eddy currents produced in the core of a transformer, large amount of energy is wasted in the form of undesirable heat. 

Minimisation of Eddy Currents: 

Eddy currents may be minimised by using laminated core of soft iron. The resistance of the laminated core increases and the eddy currents are reduced and wastage of energy is also reduced. 

Application of Eddy Currents: 

(i) Induction Furnace: 

In induction furnance, the metal to be heated is placed in a rapidly varying magnetic field produced by high frequency alternating current. Strong eddy currents are set up in the metal produce so much heat that the metal melts. This process is used in extracting a metal from its ore. The arrangement of heating the metal by means of strong induced currents is called the induction furnace. 

(ii) Induction Motor:

 The eddy currents may be used to rotate the rotor. Its principle is: When a metallic cylinder (or rotor) is placed in a rotating magnetic field, eddy currents are produced in it. According to Lenz’s law, these currents tend to reduce to relative motion between the cylinder and the field. The cylinder, therefore, begins to rotate in the direction of the field. This is the principle of induction motion.

Eddy currents. In an induction furnace, a high frequency AC is passed through a coil which surrounds the metal to be melted. The eddy currents produced in the metal heats it to high temperatures and melts it. 

Eddy currents can be minimized by slicing the conductor into pieces and laminating them so that the area for circulating currents decreases. 

Because they produce heating effect and damping effect. 

(D) Foucault currents

Eddy currents result in generation of heat (energy loss) in the cores of transformers, motors, induction coils, etc.

To minimize the eddy currents, instead of a solid metal block, cores are made of thin insulated metal strips or laminae.

Because Faraday’ law was incompatible with the law of conservation of energy. 

“The polarity of the induced emf is such that it tend to produce a current which opposes the change in magnetic flux that produced it. 

(a) Anticlockwise 

(b) Clockwise 

(c) Clockwise 

(d) Anticlockwise 

Correct answer is (a) anti-clockwise

Data : v = 5 m/s, l = 0.25 m, 

R = 10 Ω, B = 0.5T The induced current,

i = \(\cfrac{e}R\) = \(\cfrac{Blv}R\) = \(\cfrac{(0.5)(0.25)(5)}{10}\) = 0.0625 A

Since the magnetic flux into the page through the | closed conducting loop increases, the induced current in the loop must be anticlockwise. Alternatively, Fleming’s right hand rule gives the direction of induced current in the moving wire from bottom to top.

Correct answer is (b) -10 V