Answer:
Magnetic field, B = 0.016 Tesla
Explanation:
It is given that,
Velocity of electron, [tex]v=9.6\times 10^5\ m/s[/tex]
Magnetic force, [tex]F=2.6\times 10^{-15}\ N[/tex]
Charge, [tex]q=1.6\times 10^{-19}\ N[/tex]
The magnetic force is given by :
[tex]F=qvB[/tex]
[tex]B=\dfrac{F}{qv}[/tex]
[tex]B=\dfrac{2.6\times 10^{-15}}{1.6\times 10^{-19}\times 9.6\times 10^5}[/tex]
[tex]B=0.016\ T[/tex]
So, the magnitude of magnetic field is 0.016 Tesla. Hence, this is the required solution.
Answer:
The magnitude of the magnetic field strength is 16.927 mT
Explanation:
Given;
velocity of the electron, v = 9.6 x 10⁵ m/s
magnitude of the force exerted on the electron, F = 2.6 x 10⁻¹⁵ N
charge of electron, q = 1.60 x 10⁻¹⁹ C
The force exerted on the electron in magnetic field is given by charge times the vector product of velocity and magnetic field strength.
F = q(v x B)
where;
B is the magnetic field strength
[tex]B = \frac{F}{qv}[/tex]
Substitute the given values of force, charge and velocity and calculate the magnetic field strength:
[tex]B = \frac{2.6*10^{-15}}{(1.6*10^{-19})(9.6*10^5)}} \\\\B = 1.6927*10^{-2} \ T\\\\B = 16.927 \ mT[/tex]
Therefore, the magnitude of the magnetic field strength is 16.927 mT
