Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am² and 1.00 Am² 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 × 10⁻⁵ Wb/m²) (a) 3.6 × 10⁻⁵ Wb/m² (b) 2.56 × 10⁻⁴ Wb/m² (c) 3.50 × 10⁻⁴ Wb/m² (d) 5.80 × 10⁻⁴ Wb/m²

Answer: (b). O lies on the common equatorial line of both magnets. Their moments point south, so each equatorial field at O points north (opposite to M), same as BH. r = 0.1 m: B₁ + B₂ = (μ₀)/(4π)(M₁+M₂)/(r³) = 10⁻⁷ × 2.210⁻³ = 2.2 × 10⁻⁴ T B = 2.2 × 10⁻⁴ + 0.36 × 10⁻⁴ = 2.56 × 10⁻⁴ T So the answer is (b) 2.56 × 10⁻⁴ Wb/m².

Two short bar magnets of length 1 cm each have magnetic moments 1.20… | JEE Main 2013

Q24JEE Main 2013Difficult

Field of a Magnet & Magnet in a Uniform Field

Two short bar magnets of length $1\text{ cm}$ each have magnetic moments $1.20\text{ Am}^{2}$ and $1.00\text{ Am}^{2}$ 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 $20.0\text{ cm.}$ 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 $3.6 \times 10^{-5}\text{ Wb/m}^{2})$

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Answer: (b)

O lies on the common equatorial line of both magnets. Their moments point south, so each equatorial field at O points north (opposite to $M$ ), same as $B_H$ .
$r = 0.1\ \text{m}$ : $B_1 + B_2 = \frac{\mu_0}{4\pi}\frac{M_1+M_2}{r^3} = 10^{-7} \times \frac{2.2}{10^{-3}} = 2.2 \times 10^{-4}\ \text{T}$
$B = 2.2 \times 10^{-4} + 0.36 \times 10^{-4} = 2.56 \times 10^{-4}\ \text{T}$
So the answer is (b) $2.56 \times 10^{-4}\ \text{Wb/m}^2$ .

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