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class 12 physics
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?
Answers
Answer:
Given:
No. of turns in the solenoid coil, N = 15 turns/cm = 15000 turns/m
(∵ 1m = 100 cm)
∴ no. of turns in the solenoid coil per unit length, n = 15000 turns
Area of the solenoid coil = 2.0 cm2
In m2, Area will be 2 × 10 - 4 m2
Since current carried by the solenoid coil changes from 2.0 to 4.0 A in 0.1 seconds
Therefore, change in the current in the coil of solenoid = final current – initial current
di = 4.0 – 2.0 = 2.0 A
The change in time “dt” is given as ,dt = 0.1 seconds
Applying Faraday’s law, induced e.m.f can be calculated as follows:
…(1)
Where Ф is flux induced in the loop
And, flux is given as, Ф = BA
Where “B” is the magnetic field and “A” is the area of the loop.
For a solenoid, B= μ0nI
Therefore, the equation (1) becomes:
Or E = Aμ0 n (dI/dt)…(2)
∵ A, μo and n are constants
substituting the values in equation (2), we get,
⇒ E = 7.54 × 10 - 6 V
Or E = 7.54 μV
(∵ 10-6 = μ)
Hence, the induced voltage in the loop is 7.54 μV
Explanation:
Answer:
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Explanation: