# NCERT : Electromagnetic Induction

Q:3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing ?

Sol. Number of turns per unit length , n = 15 cm-1 = 1500 m-1
Area , A = 2 cm2 = 2 × 10 m-4 , μo = 4π × 10 -7
Change in current , dI = 4-2 = 2A ,
Change in time , dt = 0.1 sec
Magnetic field B = μo n I
Magnetic Flux , φ = B A = (μo n I) A
A/c to Formula , induced emf , $\displaystyle e = -\frac{d\phi}{dt}$

$\displaystyle e = -\frac{d(\mu_0 nIA)}{dt}$

$\displaystyle e = -(\mu_0 n A)\frac{dI}{dt}$

On putting all the given values we get

e = − 7.54 × 10-6

Q:4. 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 the loop is 1 cm s–1 in a direction normal to the
(a) longer side,
(b) shorter side of the loop? For how long does the induced voltage last in each case ?

Sol. using formula ,  e = B l v  and time t = length of wire / v

(a) along longer side

length , l = 8 cm = 0.08 m  , B = 0.3 T , v= 1 cm/s = 0.01 m/s

emf developed , e= B l v = 0.3 × 0.08 × 0.01 = 0.24 mV

time of emf = length of shorter arm/v = 0.02/0.01 = 2 sec.

(since , emf developed as long as loop does not get out the field , i.e. distance travelled by shorter arm )

(b)along longer side

emf developed , e= B l v =0.3 x 0.02 x 0.01 = 0.06mV

time of emf = length of longer arm/v = 0.08/0.01 =8 sec.

Q :5. 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 emf developed between the centre and the ring.

Sol . Induced emf , e = Blv

Average velocity of the rod , v = (0 + ωl)/2 = ωl/2

Induced emf , $\displaystyle e = \frac{1}{2}B\omega l^2$

On putting the values , B = 0.5 T , l = 1 m ,  ω = 400 rad s–1

e = 100 V .