Answer:
[tex]B=1.61\times 10^{-4}\ T[/tex]
Explanation:
The attached figure shows the path followed by an electron in the semicircular path from A to B.
Velocity of the electron is, [tex]v=1.42\times 10^6\ m/s[/tex]
It can be seen from the figure that the radius of thenpath, r = 5 cm or 0.05 m
The magnetic force acting on the electron is balanced by the centripetal force acting on it. It means,
[tex]Bqv=\dfrac{mv^2}{r}[/tex]
B is the magnitude of the magnetic field
[tex]B=\dfrac{mv}{rq}\\\\\text{Putting all the values}\\\\B=\dfrac{9.1\times 10^{-31}\times 1.42\times 10^6}{0.05\times 1.6\times 10^{-19}}\\\\B=1.61\times 10^{-4}\ T[/tex]
So, the magnitude of the magnetic field is [tex]1.61\times 10^{-4}\ T[/tex].
The magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B will be [tex]B=1.61\times 10^-{4]\ T[/tex]
The magnetic field is defined as when the current passes through the wire, then the magnetic field is generated around the wire in a circular pattern.
The attached figure shows the path followed by an electron in the semicircular path from A to B.
The velocity of the electron is, [tex]v=1.42\times 10^6\ \frac{m}{s}[/tex]
It can be seen from the figure that the radius of then path, r = 5 cm or 0.05 m
The magnetic force acting on the electron is balanced by the centripetal force acting on it. It means,
[tex]Bqv=\dfrac{mv^2}{r}[/tex]
B is the magnitude of the magnetic field
[tex]B=\dfrac{mv}{rq}[/tex]
[tex]B=\dfrac{9.1\times 10^{-31}\times1.42\times 10^6}{0.05\times 1.6\times 10^{-19}}[/tex]
[tex]B=1.61\times 10^{-4}\ T[/tex]
So, the magnitude of the magnetic field is [tex]B=1.61\times 10^-{4]\ T[/tex]
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