The magnetic field of the Earth at a particular location is directed vertically downward with magnitude of 40.0 μT.An electron is moving horizontally toward the west in this field with a speed of 5.2 × 10⁶ m/s.What are the direction and magnitude of the magnetic force the field exerts on the electron? (e = 1.6 × 10^-19 C) A)3.33E-17,south B)3.33E-17,north C)2.08E-20,south D)2.08E-20,north

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The magnitude of the magnetic force can be calculated using the formula $F = qvB\sin(\theta)$, where $q$ is the charge of the electron ($1.6 \times 10^{-19} C$), $v$ is the velocity of the electron ($5.2 \times 10^6 m/s$), $B$ is the magnetic field ($40.0 \mu T = 40.0 \times 10^{-6} T$), and $\theta$ is the angle between the velocity of the electron and the magnetic field, which is $90^\circ$ since the electron is moving horizontally and the field is vertical. Thus, $\sin(\theta) = \sin(90^\circ) = 1$.Substituting the values, we get:$$F = (1.6 \times 10^{-19} C) \times (5.2 \times 10^6 m/s) \times (40.0 \times 10^{-6} T) \times 1 = 3.328 \times 10^{-17} N$$The direction of the magnetic force can be determined using the right-hand rule. Since the electron is negatively charged and moving toward the west in a downward magnetic field, the force will be directed to the north.

The Magnetic Field of the Earth at a Particular Location