Two short bar magnets of length 1 cm each have magnetic moments 1.20 A m$$^2$$ and 1.00 A m$$^2$$ respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to
(Horizontal component of earth's magnetic induction is $$3.6 \times 10^{-5}$$ Wb m$$^{-2}$$)

We have two identical short magnets of length $$l = 1\;\text{cm} = 0.01\;\text{m}$$, but for the magnetic induction at a distant point only the magnetic moment is needed. Their magnetic moments are

$$M_1 = 1.20\; \text{A m}^2,$$

$$M_2 = 1.00\; \text{A m}^2.$$

The magnets are kept parallel to one another so that the line joining their centres is the common magnetic equator (broad-side line) for both of them. The distance between their centres is

$$d = 20.0\;\text{cm} = 0.20\;\text{m}.$$

Hence the mid-point $$O$$ of this line is at a distance

$$r = \dfrac{d}{2} = \dfrac{0.20\;\text{m}}{2} = 0.10\;\text{m}$$

from the centre of each magnet.

For a short bar magnet the magnetic induction at a point on its equatorial line is given by the standard formula

$$B_{\text{equator}} = \dfrac{\mu_0}{4\pi}\,\dfrac{M}{r^{3}},$$

where $$\mu_0/4\pi = 10^{-7}\;\text{T m A}^{-1}$$. The field is directed opposite to the magnetic moment vector. Since both magnets have their moments parallel, the fields produced by them at $$O$$ are along the same horizontal direction and hence add algebraically.

Magnetic induction at $$O$$ due to magnet 1:

$$B_1 = 10^{-7}\;\dfrac{M_1}{r^{3}} = 10^{-7}\;\dfrac{1.20}{(0.10)^3} = 10^{-7}\;\dfrac{1.20}{0.001} = 10^{-7}\times 1.20 \times 10^{3} = 1.20 \times 10^{-4}\;\text{T}.$$

Magnetic induction at $$O$$ due to magnet 2:

$$B_2 = 10^{-7}\;\dfrac{M_2}{r^{3}} = 10^{-7}\;\dfrac{1.00}{0.001} = 1.00 \times 10^{-4}\;\text{T}.$$

Adding these two gives the total field at $$O$$ because of the two magnets:

$$B_{\text{magnets}} = B_1 + B_2 = (1.20 + 1.00) \times 10^{-4} = 2.20 \times 10^{-4}\;\text{T}.$$

The horizontal component of the Earth’s magnetic induction is supplied as

$$B_{\text{Earth}} = 3.6 \times 10^{-5}\;\text{T} = 0.36 \times 10^{-4}\;\text{T}.$$

The equatorial fields of the magnets are directed towards geographic north (opposite to their moments), which is the same direction as the Earth’s horizontal field. Therefore the two contributions add directly. The resultant horizontal magnetic induction at $$O$$ is

$$B_{\text{resultant}} = B_{\text{magnets}} + B_{\text{Earth}} = 2.20 \times 10^{-4}\; \text{T} + 0.36 \times 10^{-4}\; \text{T} = 2.56 \times 10^{-4}\;\text{T}.$$

Expressing in the requested units of $$\text{Wb m}^{-2}$$ (which are identical to tesla), we have

$$B_{\text{resultant}} = 2.56 \times 10^{-4}\;\text{Wb m}^{-2}.$$

Hence, the correct answer is Option 4.