An electron moves with a speed of 2 times 10^ | vedclass

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(C) Given:Velocity $\vec{v} = (2 	imes 10^5 \hat{i})\,m/s$Magnetic field $\vec{B} = (\hat{i} - 4\hat{j} - 3\hat{k})\,T$Charge of electron $q = -e = -

Vedclass · Accepted Answer

$1.6 	imes 10^{-13}$

Vedclass · Answer

(C) Given:Velocity $\vec{v} = (2 	imes 10^5 \hat{i})\,m/s$Magnetic field $\vec{B} = (\hat{i} - 4\hat{j} - 3\hat{k})\,T$Charge of electron $q = -e = -1.6 	imes 10^{-19}\,C$The magnetic force is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.$\vec{v} 	imes \vec{B} = (2 	imes 10^5 \hat{i}) 	imes (\hat{i} - 4\hat{j} - 3\hat{k}) = 2 	imes 10^5 (\hat{i} 	imes \hat{i} - 4(\hat{i} 	imes \hat{j}) - 3(\hat{i} 	imes \hat{k}))$Since $\hat{i} 	imes \hat{i} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{i} 	imes \hat{k} = -\hat{j}$,we get:$\vec{v} 	imes \vec{B} = 2 	imes 10^5 (0 - 4\hat{k} + 3\hat{j}) = 2 	imes 10^5 (3\hat{j} - 4\hat{k})$.The force $\vec{F} = -1.6 	imes 10^{-19} 	imes 2 	imes 10^5 (3\hat{j} - 4\hat{k}) = -3.2 	imes 10^{-14} (3\hat{j} - 4\hat{k})$.The magnitude of the force is $F = |q| |\vec{v} 	imes \vec{B}| = 1.6 	imes 10^{-19} 	imes 2 	imes 10^5 	imes \sqrt{3^2 + (-4)^2}$.$F = 3.2 	imes 10^{-14} 	imes \sqrt{9 + 16} = 3.2 	imes 10^{-14} 	imes 5 = 16 	imes 10^{-14} = 1.6 	imes 10^{-13}\,N$.

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

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