A Helium Nucleus makes a full rotation in a Circle of Radius 0.8 meter in two seconds. The value of B at the centre of the Circle will be

Answer:

Magnetic Field is the region around a magnetic material or a moving electric charge within which the force of magnetism acts. The Magnetic field at the centre of the circle is given by,

\(B=\frac{\mu _{0}I}{2\pi r}\)………………..(1)

and

\(I = \frac{e\omega }{2\pi }\)…………………..(2)

Where

e = charge of a Helium atom (2 × 1.6 × 10-19C)

ω = Angular frequency of nth orbit

Time given is 2 seconds, so the frequency can be calculated as

\(f = \frac{1}{t}= \frac{1}{2}\)………………(3)

also

\(f = \frac{\omega }{2\pi }=\frac{1}{2}\)…..(4)

Consider equation (2)

\(I = \frac{e\omega }{2\pi }\)

where

ω/2Π = 1/2, substitute in the above equation

 

\(I = \frac{e\omega }{2\pi }= \frac{2\times 1.6\times 10^{-19}}{2}\)

 

I = 1.6 × 10-19A

 

Substitute the value of I in equation (1)

 

\(B=\frac{\mu _{0}\times 1.6\times 10^{-19}}{2\times 0.8}\)

 

B = 10-19 μo

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