An electron moves with speed 2 times 10^5 m/ | vedclass

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(C) The velocity of the electron is $\vec{v} = 2 	imes 10^5 \hat{i} \ m/s$.The magnetic field is $\vec{B} = (\hat{i} + 4\hat{j} - 3\hat{k}) \ T$.The

Vedclass · Accepted Answer

$1.6 	imes 10^{-13}$

Vedclass · Answer

(C) The velocity of the electron is $\vec{v} = 2 	imes 10^5 \hat{i} \ m/s$.The magnetic field is $\vec{B} = (\hat{i} + 4\hat{j} - 3\hat{k}) \ T$.The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.For an electron, $q = -1.6 	imes 10^{-19} \ C$.Calculating the cross product $\vec{v} 	imes \vec{B}$:$\vec{v} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 	imes 10^5 & 0 & 0 \ 1 & 4 & -3 \end{vmatrix} = \hat{i}(0) - \hat{j}(-6 	imes 10^5) + \hat{k}(8 	imes 10^5) = (6 	imes 10^5 \hat{j} + 8 	imes 10^5 \hat{k}) \ m/s \cdot T$.The magnitude of the cross product is $|\vec{v} 	imes \vec{B}| = \sqrt{(6 	imes 10^5)^2 + (8 	imes 10^5)^2} = \sqrt{36 	imes 10^{10} + 64 	imes 10^{10}} = \sqrt{100 	imes 10^{10}} = 10 	imes 10^5 = 10^6 \ m/s \cdot T$.The magnitude of the force is $F = |q| |\vec{v} 	imes \vec{B}| = (1.6 	imes 10^{-19} \ C) 	imes (10^6 \ m/s \cdot T) = 1.6 	imes 10^{-13} \ N$.

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)

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