Question

An electron of kinetic energy 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

Answer 1

The value of magnetic field B = 3.4×10^−4 T  and number of revolutions per second f = 9.6 x 10^6

Explanation:

Kinetic Energy of electron = 100 eV = 1.6×10−17 J  

12×9.1×10^−31 × V2 = 1.6×10^−17 J

V2 = 0.35×10^14  

V = 0.591×10^7 ms

Now

r = mV/qB

B = mV/qr

B = 3.3613×10^−4 T

= 3.4×10^−4 T  

No "f" cycles per second

= 1 T = 0.0951×10^6

= 9.51×10^6

Potting B = 3.361×10^−4 T

The value of "f" = 9.6 xx 10^6

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Answer 2

The applied magnetic field is $3.4 \times 10^{-4} T$. The number of electron revolutions   is $9.4 \times 10^{6}$.

Explanation:

It is given that:

An electron’s kinetic energy = 100 eV

Circle’s radius = 10 cm

12mv2 = $1.6 \times 10^{-17} J$

Here,

m is the electron mass and v is the electron speed. So,

$12 \times \mathrm{v} 2 \times 9.1 \times 10^{-31}=1.6 \times 10-17 J$

$v=0.591 \times 10^{7} \mathrm{m} / \mathrm{s}$

Now,

$r=m v e B$

$\Rightarrow B=m v e r$

$=9.1 \times 10^{-31} \times 0.591 \times 1071.6 \times 10^{-19} \times 0.1 B=3.3613 \times 10^{-4} T$

So, the magnetic field applied = $3.4 \times 10^{-4} T$

Number of revolutions of electron per second,

$f=1 T$

$T=2 \pi m B e$

$f=B e 2 \pi m=3.4 \times 10^{-4} \times 1.6 \times 10^{-192} \times 3.14 \times 9.1 \times 10^{-31}=0.094 \times 108$

$f=9.4 \times 10^{6}$