a charged particle moves at 2.5×10⁴ m/s at an angle of 25° to a magnetic field that has a field strength of…

a charged particle moves at 2.5×10⁴ m/s at an angle of 25° to a magnetic field that has a field strength of 8.1×10⁻² t. if the magnetic force is 7.5×10⁻² n, what is the magnitude of the charge? 3.7×10⁻⁵ c 4.1×10⁻⁵ c 8.8×10⁻⁵ c 1.0×10⁻⁴ c

a charged particle moves at 2.5×10⁴ m/s at an angle of 25° to a magnetic field that has a field strength of 8.1×10⁻² t. if the magnetic force is 7.5×10⁻² n, what is the magnitude of the charge? 3.7×10⁻⁵ c 4.1×10⁻⁵ c 8.8×10⁻⁵ c 1.0×10⁻⁴ c

Answer

Explanation:

Step1: Recall the magnetic - force formula

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

Step2: Rearrange the formula for $q$

Rearranging $F = qvB\sin\theta$ for $q$ gives $q=\frac{F}{vB\sin\theta}$.

Step3: Substitute the given values

We are given that $F = 7.5\times10^{-2}\text{ N}$, $v = 2.5\times10^{4}\text{ m/s}$, $B = 8.1\times10^{-2}\text{ T}$, and $\theta = 25^{\circ}$. First, find $\sin\theta=\sin(25^{\circ})\approx0.423$. Then substitute the values into the formula: [ \begin{align*} q&=\frac{7.5\times 10^{-2}}{(2.5\times 10^{4})\times(8.1\times 10^{-2})\times0.423}\ &=\frac{7.5\times 10^{-2}}{(2.5\times8.1\times0.423)\times10^{4 - 2}}\ &=\frac{7.5\times 10^{-2}}{(2.5\times8.1\times0.423)\times10^{2}}\ &=\frac{7.5\times 10^{-2}}{8.57325\times10^{2}}\ &=\frac{7.5}{8.57325}\times10^{-2 - 2}\ &\approx0.875\times10^{-4}\ &\approx8.8\times10^{-5}\text{ C} \end{align*} ]

Answer:

$8.8\times 10^{-5}\text{ C}$