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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.
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उत्तर
Area of the small flat search coil, A = 2 cm2 = 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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संबंधित प्रश्न
Prove theoretically (electromagnetic induction) `e = (dphi)/(dt)`
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