The radius of circular motion of a particle of charge q, mass m, speed v in field B (perpendicular) is:
AqB/(mv)
Bmv/(qB)
CmvB/q
Dv/(qBm)
Answer & Solution
Correct answer: B. mv/(qB)
qvB = mv²/r ⇒ r = mv/(qB).
Related questions
A storage battery of emf 8.0 V and internal resistance 0.5 $\Omega$ is charged by a 120 V The number density of free electrons in a copper wire is $8.5 \times 10^{28}\ \text{m}^{-3A heating element has resistance 100 $\Omega$ at 27.0 °C. Its temperature coefficient of rA wire of length 15 m and uniform cross-section $6.0 \times 10^{-7}\ \text{m}^2$ has a meaThe statement "$V = IR$ is itself the statement of Ohm's law" is, according to the text, iCurrent through an area of cross-section is treated as a scalar quantity in this chapter eA Wheatstone bridge has arms AB = 100 $\Omega$, BC = 10 $\Omega$, CD = 5 $\Omega$, DA = 60In a Wheatstone bridge the unknown resistance is placed in the fourth arm. Known resistanc