Electromagnetic Induction-Solved Numericals

1.
A metal rod 0.5cm long is perpendicular to a field of flux density 0.6T and moves at right angle to the field with a speed of 2 m/s. Calculate the emf induced in the rod.
ans:

l = 0.5cm

B = 0.6T

θ = 90o

v = 2m/s.

E = Blv

= 0.5 * 0.6 * 2

= 0.6V

0.6V is induced

2.

A coil of area 50cm2 is perpendicular to a uniform field of flux density 10^-3T

i)What is the flux passing through the coil/

ii) If the magnetic field drops down to 0 in 3sec, what is the value of emf induced?

ans:

i

Area (A) = 50cm2 = 5 * 10-3 m2.

B = 10-3T.

Now,

(i) Flux(θ) = B * A

= 5 * 10-6 Tm2  

ii

(ii) E = dθdt 

= 5∗10−6−03

= 1.67 * 10-6V.

3.

The mag. field of 2*10^-2T acts at right angle to a coil of area 100cm2 with 50turns. The average emf induced in the coil is o.1V when it is removed from the magnetic field in the time t.

Calculate the value of t.

ans:

B = 2 * 10-2 T

θ = 90o

A = 100cm2 = 0.01m2

n = 50 turns.

E = 0.1V

Or, θ = B * A

= 2 * 10-2 * 0.01

= 0.0002

E = N. θt$\frac{\theta }{\mathrm{t}}$

Or, 0.1 = 50 * 0.002t

So, t = 0.1 sec.

4.

Calculate the emf induced in 100cm2 circular coil having 100 turns, when the coil is taken to the mag. field of flux density 0.1T in 0.1s

ans:

A = 100cm2 = 0.01m2

N = 100 turns.

B = 0.1T

Or, t = 0.1s

Or, ϕ= BA

= 0.1 * 0.01

= 0.001.

E = N. dϕt

= 100 * 0.0010.1

= 1V.

5.

The ratio of number of turns in the primary and secondary windings of a step up transformer is 1:200. It is connected to ac mains of  200 V calculate i) Voltage developed in the secondary 

ii) The current in the secondary when the primary current is 2 A.

ans:

i

N1:N2 = 1:200

V1 = 200V

Now,

(i) or, V1V2=N1N2

Or, 200V2=1200

So, V2 = 40000V

ii

(ii) When I1 = 2A.

Or, I1I2=N2N1$\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}=\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}$

Or, 2I2=2001$\frac{2}{{\mathrm{I}}_2}=\frac{200}{1}$

So, I2 = 0.01A.

6.

An emf 3 v is produced by moving a wire 1 m long at the rate of 36km/s perpendicular to the wire and to uniform mag. field. Calculate the intensity of mag.field.

ans:

E = 3.0V

l = 1m

VR = 36km/sec

= 36000m/s.

Now,

E = Blv (θ = 90o)

Or, B = Elv

= 31∗36000

= 8.3 * 10-5T.

7.

An aeroplane with wings span 50m flying horizontally with a speed of 360Km/hr over the earth's mag. field is 0.4G. Find the potential difference between the tips of the wings.

ans:

Length of wing (l) = 50m

v = 360 km/hr.

= 360∗10003600$360\ast \frac{1000}{3600}$

= 100m/s

Earth magnetic field B = 0.4 G = 0.4* 10-4 T

PD induced (E) =?

Now,

E = BlV = 0.4 * 50 * 100 *10-4

=0. 2V.

8.

A loop of wire of area 1m² is placed perpendicular to a uniform magnetic field of 1Wbm^-2. If the field is uniformly increased to 2Wbm^-2 in a time of 10 seconds, find the induced EMF.

ans:

A = 1m², B₁ = 1Wbm^-2, B₂ = 2 Wbm^-2, t = 10s, E = ?

Initial magnetic flux, f₁ = B₁A = 1Wbm^-2 x 1m² = 1Wb

Final magnetic flux, f₂ = B₂A = 2Wbm^-2 x 1m² = 2Wb

Induced EMF

or(2-1)/10

=0.1V

9.

ans: