Question

a particle of mass 10 g and charge 80 μc moves through a uniform magnetic field, in a region where the free - fall acceleration is - 9.8ĵ m/s². the velocity of the particle is a constant 20î km/s, which is perpendicular to the magnetic field. what, then, is the magnetic field? (î + ĵ + k̂) units. attempts: 0 of 3 used save for later etextbook and media submit answer step 2

Explanation:

Step1: Recall the formula for magnetic - force and centripetal - force

The magnetic force on a charged particle is $F = qvB\sin\theta$ and the centripetal force is $F = ma$. Since the velocity is perpendicular to the magnetic field ($\theta = 90^{\circ}$, $\sin\theta=1$), and the magnetic force provides the centripetal force for the circular motion of the particle in the magnetic field. So $qvB = ma$.

Step2: Rearrange the formula to solve for $B$

We can express the magnetic field $B=\frac{ma}{qv}$. First, convert the given values to SI units. The mass $m = 10g=0.01kg$, the charge $q = 80\mu C = 80\times10^{- 6}C$, the velocity $v = 20km/s=20\times10^{3}m/s$, and the acceleration $a = 9.8m/s^{2}$.

Step3: Substitute the values into the formula

$B=\frac{0.01kg\times9.8m/s^{2}}{80\times10^{-6}C\times20\times10^{3}m/s}=\frac{0.098}{1.6}=0.06125T$.

Answer:

$0.06125T$