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A 50 cm long solenoid has 400 turns per cm. The diameter of the solenoid is 0.04 m. Find the magnetic flux linked with each turn when it carries a current of 1 A. - Physics

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प्रश्न

A 50 cm long solenoid has 400 turns per cm. The diameter of the solenoid is 0.04 m. Find the magnetic flux linked with each turn when it carries a current of 1 A.

संख्यात्मक

उत्तर

Length of the solenoid, l = 50 cm = 50 × 10-2 m

No. of turns/cm = 400

For 50 cm, No. of turns N = 400 × 50 = 20,000

Diameter of the solenoid = 0.04 m

∴ Radius of the solenoid = 0.02 m

Current passing through the solenoid = 1 A

Area of the solenoid = πr²

= 3.14 × 0.02 × 0.02 m2

Formula :-

Magnetic flux, φ = µ0 n2 AIl

n = `"N"/l`

`therefore φ = (mu_0 "N"^2 "AI")/l`

`φ = (4 xx 3.14 xx 10^-7 xx 20,000 xx 20,000 xx 3.14 xx 0.02 x 0.02 xx 1)/(50 xx 10^-2)`

`= (6310144 xx 10^-7)/(50 xx 10^-2)`

`= 126202.88 xx 10^-7 xx 10^2`

= 126202.88 × 10-5

φ = 1.262 Wb

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अध्याय 4: Electromagnetic Induction And Alternating Current - Evaluation [पृष्ठ २६२]

APPEARS IN

सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 4 Electromagnetic Induction And Alternating Current
Evaluation | Q IV. 11. | पृष्ठ २६२

संबंधित प्रश्न

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() What is the name of the effect being produced by the moving magnet?
(2) State what happens to the reading shown on the galvanometer when the magnet is moving away from the coil.
(3) The original experiment is repeated. This time the magnet is moved towards the coil at a great speed. State two changes you would notice in the reading on the galvanometer.


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Fig. shows a simple form of an A.C. generator.

(a) Name the parts labeled A and B.
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(c) Which of the output terminals is positive if the coil is rotating in the
direction shown in the diagram (anticlockwise)?
( d ) What is the position of the rotating coil when p.d. across its ends is zero? Explain why p.d. is zero when the coil is at this position .
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