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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm

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प्रश्न

It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.

संख्यात्मक
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उत्तर

Area of the small flat search coil, A = 2 cm= 2 × 10−4 m2

Number of turns on the coil, N = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 × 10−3 C

Total resistance of the coil and galvanometer, R = 0.50 Ω

Induced current in the coil,

I = `("Induced emf (e)")/"R"` ...............(1)

Induced emf is given as:

e = `-"N"("d"phi)/("dt")` ..............(2)

Where,

`"d"phi` = Charge in flux

Combining equations (1) and (2), we get

I = `-("N"("d"phi)/("dt"))/("R")`

Idt = `-"N"/"R" "d"phi` ...........(3)

Initial flux through the coil, `phi_"i"` = BA

Where,

B = Magnetic field strength

Final flux through the coil, `phi_"f"` = 0

Integrating equation (3) on both sides, we have

`int"Idt" = -"N"/"R"int_(phi_"i")^(phi_"f")"d"phi`

But total Charge Q = `int"Idt"`

∴ Q = `-"N"/"R"(phi_"f" - phi_"i") = -"N"/"R"(-phi_"i") = +("N"phi_"i")/"R"`

Q = `("NBA")/"R"`

∴ B = `("QR")/("NA")`

= `(7.5 xx 10^-3 xx 0.5)/(25 xx 2 xx 10^-4)`

= 0.75 T

Hence, the field strength of the magnet is 0.75 T.

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पाठ 6: Electromagnetic Induction - Exercise [पृष्ठ २३१]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 6 Electromagnetic Induction
Exercise | Q 6.13 | पृष्ठ २३१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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