Question

a hydrogen nucleus moves in circle of radius 1m in 1s.calculate the magnetic field produced at the centre of the circle​

Answer 1

Answer:

The current due to the revolution of the He nucleus is I=  

2π

qω

​  

=  

2

2×1.6×10  

−19

 

​  

=1.6×10  

−19

A

Magnetic field at the center B=  

2r

μ  

0

​  

I

​  

=  

2×0.8

μ  

0

​  

×1.6×10  

−19

 

​  

=10  

−19

μ  

0

​  

Explanation:

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Answer 2

Answer:

Due to orbiting of hydrogen nucleus, the magnetic field produced at the centre is equal to 3.2π×10⁻²⁷T.

Explanation:

A hydrogen nucleus has one proton. The charge on one proton is  e= 1.6×10⁻¹⁹C. The charge is moving in circle, so the magnetic field produces equal to magnetic field in a loop by a circular current.

Magnetic field produced at the centre of the circle​:

$B=\frac{\mu_{o}NI}{2r}$

where, N is the number of turns,

and r is radius of circle and I is current.

We know that  $I=\frac{q}{t}$

Here, t = 1sec and q = 1.6×10⁻¹⁹C

So, current, $I=1.6*10^{-19}A$

And. the number of turns, N=1

Given the radius of the circle, r = 1m

Substitute the value of  I, N and r in equation (1);

$B=\frac{(4\pi *10^{-7})(1)(1.6*10^{-19})}{2*(1)}$

$B= 3.2\pi *10^{26}T$

Therefore, the magnetic field produced at the centre of the circle​ is equal to 3.2π×10⁻²⁷T.