a charge of 8.4×10⁻⁴ c moves at an angle of 35° to a magnetic field that has a field strength of 6.7×10⁻³ t…

a charge of 8.4×10⁻⁴ c moves at an angle of 35° to a magnetic field that has a field strength of 6.7×10⁻³ t. if the magnetic force is 3.5×10⁻² n, how fast is the charge moving? 9.1×10⁻⁵ m/s 1.3×10⁻⁴ m/s 7.6×10³ m/s 1.1×10⁴ m/s

a charge of 8.4×10⁻⁴ c moves at an angle of 35° to a magnetic field that has a field strength of 6.7×10⁻³ t. if the magnetic force is 3.5×10⁻² n, how fast is the charge moving? 9.1×10⁻⁵ m/s 1.3×10⁻⁴ m/s 7.6×10³ m/s 1.1×10⁴ m/s

Answer

Explanation:

Step1: Recall the formula for magnetic force

The formula for the magnetic force on a moving charge is $F = qvB\sin\theta$, where $F$ is the magnetic force, $q$ is the charge, $v$ is the velocity of the charge, $B$ is the magnetic - field strength, and $\theta$ is the angle between the velocity and the magnetic field. We need to solve for $v$.

Step2: Rearrange the formula for $v$

From $F = qvB\sin\theta$, we can get $v=\frac{F}{qB\sin\theta}$.

Step3: Substitute the given values

Given $q = 8.4\times10^{-4}\ C$, $B = 6.7\times10^{-3}\ T$, $F = 3.5\times10^{-2}\ N$, and $\theta = 35^{\circ}$, and $\sin\theta=\sin35^{\circ}\approx0.574$. $v=\frac{3.5\times 10^{-2}}{(8.4\times 10^{-4})\times(6.7\times 10^{-3})\times0.574}$ First, calculate the denominator: $(8.4\times 10^{-4})\times(6.7\times 10^{-3})\times0.574=(8.4\times6.7\times0.574)\times10^{-7}\approx32.4\times10^{-7}=3.24\times10^{-6}$. Then, $v=\frac{3.5\times 10^{-2}}{3.24\times 10^{-6}}=\frac{3.5}{3.24}\times10^{4}\approx1.1\times10^{4}\ m/s$.

Answer:

$1.1\times 10^{4}\ m/s$