Question

an electron experiences a radial acceleration of 3 * 10 ^ 14 metre per second square while moving in a magnetic field along a circle of radius 0. 15 metre calculate the speed of the electron.​

Answer 1

Answer:

The speed of the electron is $2.12\times 10^{7}\text{ m/s}$

Explanation:

Centripetal acceleration of the electron

$\boxed{a=\frac{v^2}{r}}$

Therefore,

$v^2=a\times r$

Given acceleration of the electron $a=3\times 10^{14}\text{ m/s}^2$

Radius of the circular path $r=0.15\text{ m}$

Therefore, speed of the electron

$v=\sqrt{a\times r}$

$\implies v=\sqrt{3\times 10^{14}\times 0.15}$

$\implies v=\sqrt{4.5\times 10^{14}}$

$\implies v=2.12\times 10^{7}\text{ m/s}$

Therefore, the speed of the electron is $2.12\times 10^{7}\text{ m/s}$

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Answer 2

Answer:

6.7×10^14 m/s

Explanation:

a=3×10^14 m/s

r=0.15m

a=v²/r

Then v²=r×a

Now v=√3×10^14×0.15

V=√0.45×10^14

V=0.670×10^7 m/s

V=6.7×10^6 m/s

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