Question

An electron makes 3 × 105 revolutions per second in a circle of radius 0.5 angstrom. Find the magnetic field B at the centre of the circle.

Answer 1

The magnetic field B at the centre of the circle is B  = 6.03 x 10^-10

Explanation:

Magnetic field "B" = μoi/2r

No of revolutions  = 3 × 10^5 r/s

1 revolution = (1/3 × 10^5 )s

i = q/t = 1.6 x 10^-19 ÷ (1/3 × 10^5 )s

i = 3 x 1.6 x 10^-14

i = 4.8 x 10^-14 A

Now put the values in the above formula.

B = (4π x 10^-7 x 4.8 x 10^-14/2 x 0.5 x 10^-10

B  = 4.8 x 4π x 10^-11

B  = 60.3 x 10^-11

B  = 6.03 x 10^-10

Thus the magnetic field B at the centre of the circle is B  = 6.03 x 10^-10

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Answer 2

The magnetic field B at the centre of the circle is $6.69 \times 10^{-21} T$

Explanation:

Step 1:

Given data in the equation  

Electron Frequency= $3 \times 10^{5}$

The electron takes the time to complete one cycle, T =$\frac{1}{Frequency}$

Current of the  circle area , $i=\frac{q}{t}$

Loop of a Radius, r = 0.5

$A^{\circ}=0.5 \times 10^{-10} \mathrm{m}$

Step 2:

Thus the current in the loop gives the magnetic field in the centre

$B=\frac{\mu_{0} i}{2 r}$

Put the given data in the above equation  

$=\frac{\left(4 \pi \times 10^{-7} \times \frac{q}{T}\right)}{2 \times 0.5 \times 10^{-10}}$

$=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.5 \times 10^{-10} \times 3 \times 10^{5}}$

$=\frac{4 \pi \times 1.6 \times 10^{-26}}{1 \times 10^{-10} \times 3 \times 10^{5}}$

$=\frac{4 \pi \times 1.6 \times 10^{-26}}{1 \times 10^{-5} \times 3}$

$=\frac{4 \pi \times 1.6 \times 10^{-26}}{3 \times 10^{-5}}$

$=\frac{4 \times 3.14 \times 1.6 \times 10^{-21}}{3}$

$=\frac{12.56 \times 1.6 \times 10^{-21}}{3}$

$\begin{aligned}&=\frac{20.096 \times 10^{-21}}{3}\\&=6.69 \times 10^{-21} T\end{aligned}$