Question

Four like poles each of pole strength m are kept at four corners of square d find net magnetic force on the pole at any one corner

Answer 1

Four like poles each of pole strength 'm' are kept at the four corners of the square of side 'd'.find the net magnetic force on the pole at any one corner

will be :

  F¹ = F₁ + F₂ +F3

AS WE KNOW THAT

F1=F2= μ₀ m²/ 4π d²

here F1 and F2 are equal and θ= 90⁰

∴ F1+F2 = √2F [ ∵θ= 90⁰]

F3 = μ₀ m²/ 4π(√2d)²

=μ₀ m²/ 4π 2d²=F/2

∴F¹= √2F + F/2

=F (√2+1/2)

∴F¹ =[ μ₀ m²/ 4πd²] [√2+1/2]

Answer 2

➠REFER THE ROUGH DIAGRAM FROM ATTACHMENT

$\huge \orange{ \overbrace{ \red{\underbrace \color{blue} {\underbrace \color{black}{\colorbox{lime}{ {\red \: {☘ANSWER☘ }}}}}}}}$

✪QUESTION✪:-

Four like poles each of pole strength m are kept at four corners of square d find net magnetic force on the pole at any one corner

✪ANSWER✪:-

➥As we know like charges repel ,the force will move away from the charges due to repulsion

➥Like this, F1 and F2 moves away from the poles

➥And the electric force will be

F= μ₀ m²/ 4π d²

➥since all poles have same pole strength ,the forces also will be same

➥Therefore ,F1 =F2

➥since F1 and F2 intersected to each other and forms 90°,the resultant force (F3) from this ,can be calculated using parallelogram law of vectors ( as forces are vector quantities)

According to parallelogram of vectors

➠$F3 = \sqrt{F1 + F2 + 2F1F2cosθ}$

here ,F1 = F2 = F= μ₀ m²/ 4π d²

➠$F3 = \sqrt{ {F}^{2} + {F}^{2} + 2FFcos90}$

➠$F3 = \sqrt{2 {F}^{2} + 0}$

since cos90° = 0

therefore

➠$F3 = \sqrt{2} F$

➥then through diagonal ,another force(F4)moves away from the poles and added with F3 as they are moving in same direction .

➥F4 = μ₀ m²/ 4π (√2d)²

(since it moves through diagonal ,the distance between the poles will be √2d )

➥F4 = μ₀ m²/ 4πd²(2)

➥F4 = F/2 (F= μ₀ m²/ 4π d²)

so total resultant will be

➠F' = F3+F4

• F3= √2F
• F4= F/2

➠F' = √2F + F/2

➠F' = (2√2F +F )/2

➠F' = (2√2 +1 ) F/2