Example 1

The magnitude of the induced emf

At t = 2.0 s, ε = 10 x 2 + 6.0 = 26 mV

Example 2

The magnitude of the induced emf

volt

Example 3

The magnitude of the induced emf

Example 4

The magnetic flux passing through the loop is given by **Φ _{B} = BA cosθ = B_{0} sin ωt x πr^{2} x cos 0^{0} = B_{0} πr^{2} sin ωt**

The magnitude of the induced emf,

The magnitude of the induced current,

Example 5

The reading of the millivoltmeter = motional emf produced across the axle = Bvl

Speed v = 135 km/h = 135 x 5/18 m/s = 37.5 m/s

Magnetic field B = 0.2 G = 0.2 x 10** ^{- 4}** T

Length l = 1 m

Therefore, the reading of the millivoltmeter = 0.2 x 10

Example 6

Wing span l = 50 m

Speed v = 200 m/s

Horizontal component of earth's field BH = 0.3 gauss = 0.3 x 10^{- }^{4} T

Vertical component of earth's field BV = 0.5 gauss = 0.5 x 10^{- }^{4} T

An aeroplane flying horizontally cuts only the vertical component of the earth's field. So, the potential difference
between the wing tips of the aeroplane is

ε = B_{v}vl = 0.5 x 10^{- }^{4} x 200 x 50 = 0.5 V

Example 7

The network of resistors forms a balanced Wheatstone bridge and hence resistor of 3 ohm in the middle can
be neglected. Remaining resistors have an equivalent resistance = 3 ohm. Total resistance = 3 + 1 = 4 ohm.

Induced emf, Bvl = IR

Or 2 x v x 0.10 = 1 x 10^{- }^{3} x 4

This gives v = 2 cm/sec.

Example 8

The initial magnetic flux enclosed by the rod and the rails (at time t = 0) = BA = B_{0}x_{0}l.

The induced emf is given by

As a result of the induced emf a current will flow through the rod with a magnitude equal to

The direction of the current is along the wire, and therefore perpendicular to the magnetic field. The force exerted by the magnetic field on the rod is given by

The acceleration of the rod is therefore given by

Example 9

Emf induced between the rim and the axle is

Example 10

Length of rod l = 1 m

Speed v = 72 km/h = 20 m/s

Angle of dip δ = 30^{0}

A rod falling vertically along east-west direction cuts only the horizontal component of the earth's field.

Vertical component of earth's field B_{v} = 1/2 √3 gauss = (1/2 √3) x 10^{- }^{4} T

Horizontal component of earth's field B_{H} = B_{v} / tan δ = (1/2 √3)/(1 / √3) gauss = 0.5 x 10^{-}^{4} T

So, the potential difference between the ends of is

ε = B_{H}vl = 0.5 x 10^{- }^{4} x 20 x 1 = 10^{- }^{3} V

CBSE Electromagnetic Induction ( With Hint / Solution)

Class XII (By Mr. Ashis Kumar Satapathy)

email - [email protected]

Electromagnetic Induction