A bar magnet of length 3 cm has points A and B along its axis at distances of 24 cm and 48 cm on the opposite sides. Ratio of magnetic fields at these points will be

Question

A bar magnet of length 3 cm has points A and B along its axis at distances of 24 cm and 48 cm on the opposite sides. Ratio of magnetic fields at these points will be (A) 8 (B) (1/2√2) (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Magnetic field due to a bar magnet at a distance r from the centre of magnet on axial position =(μ0/4π) ⋅ (2M/r3) ∴ (B1/B2)= Bigg((r2/r1) Bigg)3= Bigg((48/24) Bigg)3 =(8/1)=8

Q. A bar magnet of length $3 c m$ has points $A$ and $B$ along its axis at distances of $24 c m$ and $48 c m$ on the opposite sides. Ratio of magnetic fields at these points will be

8

$\frac{1}{2 2}$

3

4

Magnetic field due to a bar magnet at a distance r from the centre of magnet on axial position
$= \frac{μ _{0}}{4 π} \cdot \frac{2 M}{r ^{3}}$
$∴ \frac{B _{1}}{B _{2}} = (\frac{r _{2}}{r _{1}})^{3} = (\frac{48}{24})^{3}$
$= \frac{8}{1} = 8$

Q. A bar magnet of length 3cm has points A and B along its axis at distances of 24cm and 48cm on the opposite sides. Ratio of magnetic fields at these points will be

8

22​1​

3

4

Magnetic field due to a bar magnet at a distance r from the centre of magnet on axial position =4πμ0​​⋅r32M​ ∴B2​B1​​=(r1​r2​​)3=(2448​)3 =18​=8

Q. A bar magnet of length $3 c m$ has points $A$ and $B$ along its axis at distances of $24 c m$ and $48 c m$ on the opposite sides. Ratio of magnetic fields at these points will be

$\frac{1}{2 2}$

Magnetic field due to a bar magnet at a distance r from the centre of magnet on axial position
$= \frac{μ _{0}}{4 π} \cdot \frac{2 M}{r ^{3}}$
$∴ \frac{B _{1}}{B _{2}} = (\frac{r _{2}}{r _{1}})^{3} = (\frac{48}{24})^{3}$
$= \frac{8}{1} = 8$