Question

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.

Answer 1

Here , 

Frequency of oscillations, `V_1 = 40` oscillations/min

Earth's horizontal magnetic field, B_H = 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
B_H = 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"`