A proton is projected with velocity overright | vedclass

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(A) The Lorentz force is given by $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{V} 	imes \overrightarrow{B})$.Given: $q = 1.6 	imes 1

Vedclass · Accepted Answer

$1400$

Vedclass · Answer

(A) The Lorentz force is given by $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{V} 	imes \overrightarrow{B})$.Given: $q = 1.6 	imes 10^{-19} \; C$,$m = 1.67 	imes 10^{-27} \; kg$ (approximated to $1.6 	imes 10^{-27} \; kg$ for calculation),$\overrightarrow{V} = 2 \hat{i} \; m/s$,$\overrightarrow{E} = 10 	imes 10^{-6} \hat{i} \; V/m$,$\overrightarrow{B} = (\hat{i} + 3 \hat{j} + 4 \hat{k}) 	imes 10^{-6} \; T$.First,calculate the cross product: $\overrightarrow{V} 	imes \overrightarrow{B} = (2 \hat{i}) 	imes (\hat{i} + 3 \hat{j} + 4 \hat{k}) 	imes 10^{-6} = (6 \hat{k} - 8 \hat{j}) 	imes 10^{-6} \; V/m$.Now,$\overrightarrow{F} = q [10 \hat{i} 	imes 10^{-6} + (6 \hat{k} - 8 \hat{j}) 	imes 10^{-6}] = q 	imes 10^{-6} [10 \hat{i} - 8 \hat{j} + 6 \hat{k}] \; N$.Acceleration $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{1.6 	imes 10^{-19} 	imes 10^{-6} [10 \hat{i} - 8 \hat{j} + 6 \hat{k}]}{1.6 	imes 10^{-27}} = 10^2 [10 \hat{i} - 8 \hat{j} + 6 \hat{k}] = [1000 \hat{i} - 800 \hat{j} + 600 \hat{k}] \; m/s^2$.The magnitude of acceleration is $a = \sqrt{1000^2 + (-800)^2 + 600^2} = \sqrt{1000000 + 640000 + 360000} = \sqrt{2000000} \approx 1414 \; m/s^2$. Given the options,$1400$ is the intended answer.

$A$ proton is projected with velocity $\overrightarrow{V} = 2 \hat{i} \; m/s$ in a region where magnetic field $\overrightarrow{B} = (\hat{i} + 3 \hat{j} + 4 \hat{k}) \; \mu T$ and electric field $\overrightarrow{E} = 10 \hat{i} \; \mu V/m$. Find the net acceleration of the proton (in $m/s^2$).

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