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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 - Physics

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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

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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अध्याय 6: Electromagnetic Induction - Exercise [पृष्ठ २३१]

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एनसीईआरटी Physics [English] Class 12
अध्याय 6 Electromagnetic Induction
Exercise | Q 13 | पृष्ठ २३१
एनसीईआरटी Physics [English] Class 12
अध्याय 6 Electromagnetic Induction
Exercise | Q 6.13 | पृष्ठ २३१

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