Predict the direction of induced current in the situations descri
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A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of loop is 1 cm s–1 in a direction normal to the (i) longer side (ii) shorter side of the loop? For how long does the induced voltage last in each case?

Given,
Length of loop, l = 8 cm = 8 x 10–2 m
Breadth of loop, b = 2 cm = 2 x 10–2 m
Strength of magnetic field, B = 0.3 T
Velocity of loop, v = 1 cm/sec = 10–2 m/sec 

Let the field be perpendicular to the plane of the paper directed inwards. 

(i)Then, the magnitude of induced emf is,
           
             = 0.3 × 8 × 10-2 × 10-2= 2.4 × 10-4V 

Time for which induced e.m.f. will last is equal to the time taken by the coil to move outside the field
            t = distance travelledvelcoity = 2×10-210-2m = 2 sec. 

(ii) The conductor is moving outside the field normal to the shorter side.
b = 2 x 10–2 m 

∴ The magnitude of induced emf is
                     ε = B.b.ν
                       = 0.3 × 2 × 10-2 × 10-2= 0.6 × 10-4V 

Time, t = distancevelocity =8 × 10-210-2 = 8 sec.

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Use Lenz’s law to determine the direction of induced current in the situation described by Fig:
(a)    A wire of irregular shape turning into a circular shape;
(b)    A circular loop being deformed into a narrow straight wire.



(a) When a wire of irregular shape turns into a circular loop, area of the loop tends to increase. Therefore, magnetic flux linked with the loop increases. According to Lenz’s law, the direction of induced current must oppose the change in magnetic field, for which induced current should flow along adcba (anticlockwise). 

(b) In this case, the magnetic flux tends to decrease. Therefore, induced current must support the magnetic field, for which induced current should flow along adcba (anticlockwise).

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Predict the direction of induced current in the situations described by the following figs.(a) to(f).


 


(a) South pole develops at end q and the induced current should flow clockwise. Therefore, induced current in the coil flows from qr to pq. 

(b) Coil pq in this case would develop S-pole at q and coil XY would also develop S pole at X. Therefore, induced current in coil pq will be from q to p and induced current in the coil XY will be from Y to X. 

(c) North pole is moving away from the coil hence, induced current in the right loop will be along XYZ. 

(d) Induced current in the left loop will be along ZYX as seen from front. 

(e) Induced current in the right coil is from X to Y. 

(f) Since magnetic lines of force lie in the plane of the loop, no current is induced.

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A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the e.m.f. developed between the centre and the ring.

Here,
Length of the rod, l = 1 m
Angular frequency of roation, ω = 400 s
–1
Uniform magnetic field, B = 0.5 T
e.m.f  =?

Note that linear velocity of one end of rod is zero and linear velocity of other end is (l ω).
Therefore,
Average linear velocity
 v = 0+2 = l ω/2. (  ν = ) 

Hence, 

           e = Blv    = Bll ω2  = Bl2ω2   = 0.5 × 12 × 4002  = 100 V
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A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0A in 0.1 s, what is the induced emf in the loop while the current is changing?

Given, a solenoid.
Number of turns per unit length,n = 15 turns/cm = 1500 turns/m
Area of the small loop, A = 2 cm
2 = 2 x 10–4 m2
Initial current, I1 = 2A
Final current, I
2 = 4A
Δt = 0.1s

The magnetic field associated with current I1,
                     B1 = μ0 n I1 

The magnetic field associated with current I2,
                     B2 = μ0 n I2 

The change in flux assosciated with change in current in solenoid, 

ϕ = (B2-B1) A         = 4π×10-7×1500 × (4-2) × 2 × 10-4
     = 7.6 × 10-7 weber

The induced EMF while current is changing,   
                 E = ϕt       = 7.6 × 10-70.1      = 7.6 × 10-6V
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