Question
Question: A helium nucleus is moving in a circular path of radius \(0.8m\). If it takes \(2\text{ }sec\) to co...
A helium nucleus is moving in a circular path of radius 0.8m. If it takes 2 sec to complete one revolution, then what is the magnetic field produced at the centre of the circle?
A. μ0×10−19T
B. μ010−19T
C. 2×10−19T
D. μ02×10−19T
Solution
A moving charge produces a magnetic field. Helium nucleus has two protons. Charge moving in a circle is equivalent to current in a circular loop. Use the formula of magnetic field due to circular loop to obtain the magnetic field produced due to helium nucleus.
Formula used:
Magnetic field at centre of a current carrying loop, B=2rμ0NI
Complete step by step answer:
A helium nucleus has two protons. Charge on each proton is e=1.6×10−19C. Therefore charge on helium nucleus is 2e. Charge moving in a circle produces a magnetic field which is equivalent to magnetic field produced by a circular current carrying loop.
Magnitude of magnetic field at centre of a current carrying loop is given by,
B=2rμ0NI
Where N is the number of turns, I is the current and r is the radius of the circle.
Since, the Helium nucleus takes 2 sec to complete one revolution. This means charge per unit time i.e. current due to helium nucleus is
I=q/t=22e=e=1.6×10−19A
Number of turns, N=1
Helium nucleus is moving in a circular path of radius r=0.8m.
Therefore, substituting these values, we get
B=2×0.8μ0×1×(1.6×10−19)=μ0×10−19T
The magnetic field produced due to oriting of helium nucleus at centre is μ0×10−19T which is option A.
Hence, A is the correct option.
Note:
As static charge produces electric field, the charge in motion produces magnetic field.
Magnetic field is not produced if charge is not moving but electric field is still generated due to a charge. The circular motion of the nucleus of an atom produces a magnetic field as does electrical current flowing through a wire.