Exercise 15
Wing span l = 25 m
Speed v = 1800 km/h = 500 m/s
An aeroplane flying horizontally cuts only the vertical component of the earth's field.
Dip angle, δ = 300
Vertical component of earth's field BV = B sin δ = 5 10- 4 x sin 300 = 2.5 x 10- 4
So, the potential difference between the wing tips of the aeroplane is ε = Bvvl = 2.5 x 10- 4 x 500 x 25 = 3.1 V
Exercise 16
Magnetic filed inside the solenoid,
Flux through the solenoid,
Magnitude of emf induced,
Exercise 17
Magnetic flux through the coil Φ = NBA = 500 x 0.2 x 4 x 10- 4 Wb = 0.04 Wb
Change in flux when the coil is rotated through 1800 is
ΔΦ = 0.04 – ( - 0.04) = 0.08 Wb
Exercise 18
Exercise 19
The solution of this question is based on the fact that just after closing the switch (i.e. t = 0), the induced emf or back emf is maximum and it does not allow any current to pass through the inductor. Therefore, inductor behaves as an open circuit at t = 0.
The circuit will look like as shown in figure 21 (a). But long after closing the switch (i.e. t = ∞), the induced emf or back emf becomes zero and it allows the current to pass through the inductor. Therefore, inductor behaves as a short circuit at t =∞. The circuit will look like as shown in figure 21 (b).
Exercise 20
CBSE Electromagnetic Induction ( With Hint / Solution)
Class XII (By Mr. Ashis Kumar Satapathy)
email - [email protected]
Electromagnetic Induction