मराठी

A Small Flat Search Coil of Area 5cm2 with 140 Closely Wound Turns is Placed Between the Poles of a Powerful Magnet Producing Magnetic Field 0.09t and Then Quickly Removed Out of the Field Region. - Physics

Advertisements
Advertisements

प्रश्न

A small flat search coil of area 5cm2 with 140 closely wound turns is placed between the poles of a  powerful magnet producing magnetic field 0.09T and then quickly removed out of the field region. Calculate:

(a) Change of magnetic flux through the coil, and

(b) emf induced in the coil.

संख्यात्मक
Advertisements

उत्तर १

a = 5cm2

= 5 × 10-4 m

n = 140 

(a) B = 0.09T

`Φ = bar(B) . bar(A) = BA cos theta = 0.09 xx 5 xx10^(-4) xx 1`

 = 0.45 × 10-4 W b

(b) `e = n (d phi )/(dt)`

= `140 xx (0.45 xx 10^(-4))/dt`

but dt is very small
dt → 0

`therefore -> ∞` 

shaalaa.com

उत्तर २

Number of turns in the coil, N = 140
Area of cross-section of the coil = 5 cm2
The magnetic field, B = 0.09 T 

(a) Flux linked with the coil, Φ = NBA cos θ

θ = 0° , Φi = NBA

⇒ Φi  = 140 × 0.09 × 5 × 10-4

⇒ Φi = 6.3 × 10-3 Weber

When the coil is quickly removed, the flux linkage becomes zero,Φf = 0

Hence change in magnetic flux through the coil, 

ΔΦ = Φfi = (0 - 6.3 × 10-3) Weber

ΔΦ = - 6.3 × 10-3 Weber

(b) Let the coil is removed in a very short period of time-interval, dt = 1s, then induced emf, 

`"e" = -("d"phi)/("dt") = - ((-6.3xx10^-3)/1) = 6.3 xx 10^-3 "V"`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2018-2019 (March) 55/3/3
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×