Explanation:
It is given that,
Potential difference, [tex]V=71600\ V[/tex]
Magnetic field, [tex]B=0.345\ T[/tex]
The magnetic field is perpendicular to the direction of motion of electron i.e. the angle between velocity and magnetic filed is 90. We need to calculate the magnitude of the force on the electron due to the magnetic field.
[tex]F=qvB[/tex]
Using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=eV[/tex]
[tex]v=\sqrt{\dfrac{2eV}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 71600}{9.1\times 10^{-31}}}[/tex]
[tex]v=1.58\times 10^8\ m/s[/tex]
Equation (1) becomes :
[tex]F=1.6\times 10^{-19}\times 1.58\times 10^8\times 0.345[/tex]
[tex]F=8.72\times 10^{-12}\ N[/tex]
So, the force acting on the electron due to this magnetic field is [tex]8.72\times 10^{-12}\ N[/tex]. Hence, this is the required solution.
The magnitude of the magnetic force is 8.72 x 10⁻¹² N.
The velocity of the electron is determined from the principle of conservation of energy.
[tex]K.E= eV\\\\\frac{1}{2} mv^2= eV\\\\mv^2 = 2eV\\\\v^2 = \frac{2eV}{m} \\\\v = \sqrt{\frac{2eV}{m}} \\\\v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 71600}{9.11 \times 10^{-31}}}\\\\v = 1.58 \times 10^{8} \ m/s[/tex]
The magnitude of the magnetic force is calculated as follows;
F = qvB
F = 1.6 x 10⁻¹⁹ x 1.58 x 10⁸ x 0.345
F = 8.72 x 10⁻¹² N
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