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3 years ago

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Simora [160]3 years ago

6 0

5 1 4 are the answers

kozerog [31]3 years ago

4 0

Answer:

The answers are 1.) has magnitude 4.) equals velocity times acceleration and lastly 5.) is measured in m/s2 Hope this helps :D

Explanation:

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It is given that,

Kinetic energy of the electron, $E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J$

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The magnetic field in north direction, $B_y=19911.5\ nT$

The magnetic field in west direction, $B_x=-3257.1\ nT$

The magnetic field in vertical direction, $B_z=48381.8 \ nT$

Magnetic field, $B=(-3257.1i+19911.5j+48381.8k)\ nT$

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$v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}$

$v=5.92\times 10^7 i\ m/s$ (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

$F=q(v\times B)$

$F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})$

Since, $i\times j=k\ \\j\times k=i\\k\times i=j$

And, $i\times i=j\times j=k\times k=0$

$F=1.6\times 10^{-19}\times [1178 k-2864.20j]$

$|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}$

$F=4.95\times 10^{-16}\ N$

(b) Let a is the acceleration of the electron. It can be calculated as :

$a=\dfrac{F}{m}$

$a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}$

$a=5.43\times 10^{14}\ m/s^2$

Hence, this is the required solution.

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