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प्रश्न
A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is 25 μT. Another short magnet of magnetic moment 1.6 A m2 is placed 20 cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole (a) towards north and (b) towards south.
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उत्तर
Here ,
Frequency of oscillations, `V_1 = 40` oscillations/min
Earth's horizontal magnetic field, BH = 25 μT
Magnetic moment of the second magnet, M = 1.6 A-m2
Distance at which another short magnet is placed, d = 20 cm = 0.2 m
(a) For the north pole of the short magnet facing the north, frequency `(V_1)` is given by
`V_1 = 1/(2pi)sqrt((MB_H)/I)`
Here,
M = Magnetic moment of the magnet
I = Moment of inertia
BH = Horizontal component of the magnetic field
Now, let B be the magnetic field due to the short magnet.
When the north pole of the second magnet faces the north pole of the first magnet, the effective magnetic field `(B_(eff))` is given by
`B_(effective) = B_H - B`
The new frequency of oscillations (`V_2`) on placing the second magnet is given by `V_2 = 1/(2pi) sqrt((M(B_H-B))/I`
The magnetic field produced by the short magnet (`B`) is given by
`B = (u_0)/(4pi) m/d^3`
⇒ `B = (10^-7 xx 1.6)/(8 xx 10^-3) = 20 "uT"`
Since the frequency is proportional to the magnetic field,
`V_1/V_2 = sqrt (B_H/(B_H - B)`
⇒ `40/V_2 = sqrt(25/5)`
⇒ `40/V_2 = sqrt(5)`
⇒ `V_2 = 40/sqrt(5) = 17.88`
= 18 oscillations/min
(b) For the north pole facing the south,
`V_1 = 1/(2pi) sqrt((MB_H)/I)`
⇒ `V_2 = 1/(2pi) sqrt((M(B+B_H))/I)`
⇒ `V_1/V_2 = sqrt(B_H/(B+B_H))`
⇒ `40/V_2 = sqrt(25/45)`
⇒ `V_2 = 40/sqrt("25/45") = 54 "oscillation/min"`
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