A 1 mutext{C} charge is moving with velocity | vedclass

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(B) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.First,calculate the cross product $\vec{v} 	imes \vec{B}$

Vedclass · Accepted Answer

$171$

Vedclass · Answer

(B) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.First,calculate the cross product $\vec{v} 	imes \vec{B}$:$\vec{v} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -5 \end{vmatrix}$$= \hat{i}((-2)(-5) - (3)(3)) - \hat{j}((1)(-5) - (3)(2)) + \hat{k}((1)(3) - (-2)(2))$$= \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$$= 1\hat{i} + 11\hat{j} + 7\hat{k}$.The magnitude of the cross product is $|\vec{v} 	imes \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$.Given $q = 1 \mu	ext{C} = 10^{-6} 	ext{ C}$,the force magnitude is $F = |q| |\vec{v} 	imes \vec{B}| = 10^{-6} 	imes \sqrt{171} 	ext{ N}$.Comparing this with $\sqrt{\alpha} 	imes 10^{-6} 	ext{ N}$,we get $\alpha = 171$.

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