A short magnet of moment 6.75 Am 2 produces a neutral point on its axis. If horizontal component of earth's magnetic field is 5 × 10-5 Wb / m 2, then the distance of the neutral point should be

Question

A short magnet of moment 6.75 Am 2 produces a neutral point on its axis. If horizontal component of earth's magnetic field is 5 × 10-5 Wb / m 2, then the distance of the neutral point should be (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm

Tardigrade Admin · Accepted Answer

At neutral point, | textMagnetic field due to magnet|= | textMagnetic field due to earth| (μ0/4 π) ⋅ (2 M/d3)=5 × 10-5 ⇒ 10-7 × (2 × 6.75/d3)=5 × 10-5 d=0.3 m =30 cm

Q. A short magnet of moment $6.75 A m^{2}$ produces a neutral point on its axis. If horizontal component of earth's magnetic field is $5 \times 1 0^{- 5} Wb / m^{2}$ , then the distance of the neutral point should be

10 cm

20 cm

30 cm

40 cm

At neutral point,
$∣ Magnetic field due to magnet ∣ = ∣ Magnetic field due to earth ∣$
$\frac{μ _{0}}{4 π} \cdot \frac{2 M}{d ^{3}} = 5 \times 1 0^{- 5}$
$\Rightarrow 1 0^{- 7} \times \frac{2 \times 6.75}{d ^{3}} = 5 \times 1 0^{- 5}$
$d = 0.3 m = 30 c m$

Q. A short magnet of moment 6.75Am2 produces a neutral point on its axis. If horizontal component of earth's magnetic field is 5×10−5Wb/m2, then the distance of the neutral point should be