QuestionAugust 5, 2025

What is the force exerted on a moving charge of -2.0mu C at a 20^circ angle through a magnetic field of 3.0times 10^-4 T with a velocity of 5.0times 10^6m/s 3.0times 10^-3 1.0times 10^-3 1.0times 10^3

What is the force exerted on a moving charge of -2.0mu C at a 20^circ angle through a magnetic field of 3.0times 10^-4 T with a velocity of 5.0times 10^6m/s 3.0times 10^-3 1.0times 10^-3 1.0times 10^3
What is the force exerted on a moving charge of -2.0mu C at a 20^circ angle through a magnetic field of 3.0times 10^-4 T with a velocity of 5.0times 10^6m/s 3.0times 10^-3 1.0times 10^-3 1.0times 10^3

Solution
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Answer

-1.026 \times 10^{-3} \, N Explanation 1. Identify the formula The force on a moving charge in a magnetic field is given by **F = qvB\sin(\theta)**. 2. Substitute values q = -2.0 \times 10^{-6} \, C, v = 5.0 \times 10^{6} \, m/s, B = 3.0 \times 10^{-4} \, T, \theta = 20^\circ. 3. Calculate the sine of the angle \sin(20^\circ) \approx 0.342. 4. Compute the force F = (-2.0 \times 10^{-6}) \times (5.0 \times 10^{6}) \times (3.0 \times 10^{-4}) \times 0.342.

Explanation

1. Identify the formula<br /> The force on a moving charge in a magnetic field is given by **$F = qvB\sin(\theta)$**.<br />2. Substitute values<br /> $q = -2.0 \times 10^{-6} \, C$, $v = 5.0 \times 10^{6} \, m/s$, $B = 3.0 \times 10^{-4} \, T$, $\theta = 20^\circ$.<br />3. Calculate the sine of the angle<br /> $\sin(20^\circ) \approx 0.342$.<br />4. Compute the force<br /> $F = (-2.0 \times 10^{-6}) \times (5.0 \times 10^{6}) \times (3.0 \times 10^{-4}) \times 0.342$.
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