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Question Number 213221 by mr W last updated on 01/Nov/24

Commented by mr W last updated on 01/Nov/24

$Q 212936$
Commented by ajfour last updated on 01/Nov/24
https://youtu.be/tB8F0OnMlAU?si=fmSM55gGzwSidaU-
Commented by aleks041103 last updated on 01/Nov/24

$I s a w t h e s o l u t i o n a n d i t s e e m s y o u$ $n e g l e c t t h e a c t i o n o f g r a v i t y o n t h e$ $r i g h t h a l f p l a n e .$ $I f y o u a c o u n t f o r g t h e r e t o o, t h e p r o b l e m$ $b e c o m e s r e a l l y d i f f i c u l t, s i n c e y o u w o u l d$ $h a v e t o s o l v e a s y s t e m o f e q u a t i o n s$ $w h i c h a r e t r a n s c e n d e n t a l .$
	Answered by mr W last updated on 01/Nov/24

	$v_{1} = u \cos α$ $h = \frac{u^{2} \sin^{2} α}{2 g}$ $a = \frac{2 h}{\tan α} = \frac{u^{2} \sin^{2} α}{g \tan α}$ $v_{2 x} = v_{1} = u \cos α$ $r = \frac{{m v}_{1}}{q B} = \frac{m u \cos α}{g B}$ $h - b = \frac{2 m u \cos α}{q B} + \frac{g}{2} {(\frac{π m}{q B})}^{2}$ $v_{2 y} = \sqrt{2 g (h - b)} = \sqrt{2 g [\frac{2 m u \cos α}{q B} + \frac{g}{2} {(\frac{π m}{q B})}^{2}]}$ $b = v_{2 y} \times \frac{a}{u \cos α} + \frac{{g a}^{2}}{2 u^{2} \cos^{2} α}$ $\frac{u^{2} \sin^{2} α}{2 g} - \frac{2 m u \cos α}{q B} - \frac{g}{2} {(\frac{π m}{q B})}^{2} = \sqrt{2 g [\frac{2 m u \cos α}{q B} + \frac{g}{2} {(\frac{π m}{q B})}^{2}]} \times \frac{\frac{u^{2} \sin^{2} α}{2 g}}{u \cos α} + \frac{g {(\frac{u^{2} \sin^{2} α}{2 g})}^{2}}{2 u^{2} \cos^{2} α}$ $\frac{\sin^{2} α}{2} - \frac{2 m g \cos α}{u q B} - \frac{1}{2} {(\frac{π m g}{u q B})}^{2} = \sqrt{\frac{4 m g \cos α}{u q B} + {(\frac{π m g}{u q B})}^{2}} \times \frac{\sin^{2} α}{2 \cos α} + \frac{\sin^{4} α}{8 \cos^{2} α}$ $l e t λ = \frac{m g}{u q B}$ $(4 λ \cos α + π^{2} λ^{2}) + \sin α \tan α \sqrt{4 λ \cos α + π^{2} λ^{2}} + (\frac{\tan^{2} α}{4} - 1) \sin^{2} α = 0$ $\sqrt{4 λ \cos α + π^{2} λ^{2}} = \frac{- \sin α \tan α + \cos α}{2} = \frac{\cos 2 α}{2 \cos α}$ $π^{2} λ^{2} + 4 λ \cos α - \frac{\cos^{2} 2 α}{4 \cos^{2} α} = 0$ $\Rightarrow λ = \frac{m g}{u q B} = \frac{- 4 \cos^{2} α + \sqrt{16 \cos^{4} α + π^{2} \cos^{2} 2 α}}{2 π^{2} \cos α}$
	Commented by ajfour last updated on 01/Nov/24
	$yes, v nice, overall i followed and agree sir. Lets say cos α=s B=((2mg)/(qu)){((4s^2 +(√(16s^4 +π^2 (2s^2 −1)^2 )))/((1/s)(2s^2 −1)^2 ))}$
	$y e s, v n i c e, o v e r a l l i f o l l o w e d a n d a g r e e$ $s i r . L e t s s a y \cos α = s$ $B = \frac{2 m g}{q u} {\frac{4 s^{2} + \sqrt{16 s^{4} + π^{2} {(2 s^{2} - 1)}^{2}}}{\frac{1}{s} {(2 s^{2} - 1)}^{2}}}$
	Commented by mr W last updated on 01/Nov/24

	$t h a n k s f o r c h e c k i n g s i r!$
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