Question

An alpha particle (2e) moves along a circular path of radius 0.1A with a uniform speed of 2*10^6 m/s.Calculate the magnetic field produced at the centre of the circular path.

Answer 1

time period(T)=2*pi*r/v

equivalent current (i) =2e/T

k= 4pi*10^-7
magnetic field =ki/2r

put the values to get the answer

Answer 2

The magnetic field produced at the centre of the circular path is 6.34 A/m

Given,

An alpha Particle (2e) moves along a circular path.

Radius = 0.1 A

Uniform speed = $2* 10^{6} m/s$

To Find,

Find the magnetic field produced at the centre of the circular path.

Solution,

According to the question,

speed of alpha particle = $2* 10^{6} m/s$

constant Value of Unit Charge (e) = $1.6 * 10^{-19} C$

Radius = 1A° = $1* 10^{-10} m$

As magnetic field at centre of circuit loop B = $\frac{u_{0} * I}{2R}$ .......(1)

$[Where, u_{0} = 4*10^{-7} T m/A]$

Current , I = $\frac{2e}{T}$ ......(2)

Time Period, T = $\frac{2πR}{V}$

$= > \frac{2π*1*10^{-10}}{2*10^{-6} } \\\\= > 3.14 * 10^{-16} s$

Now, putting the value of T in equation 2 to find the value of I

so, I = $\frac{3.2* 10^{-19} }{3.14* 10^{-16} }$

$= > 1.01 * 10^{-3} A$

Therefore, B = $\frac{4π* 10^{-7}* 1.01* 10^{-3} }{2*10^{-10} }$

so, value of B = 6.34 A/m

Hence, the magnetic field produced at the centre of the circular path is 6.34 A/m

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