Question
A deuteron of kinetic energy $50 \mathrm{keV}$ is describing a circular orbit of radius $0.5 \mathrm{~m}$ in a plane perpendicular to the magnetic field $\vec{B}$. The kinetic energy of the proton that describes a circular orbit of radius $0.5 \mathrm{~m}$ in the same plane with the same field $\vec{B}$ is(a) $5 \mathrm{keV}$(b) $10 \mathrm{keV}$(c) $50 \mathrm{keV}$(d) $100 \mathrm{keV}$
Step 1
In this case, the particle is moving in a circle, so the acceleration is centripetal, and the force is directed towards the center of the circle. This gives us $qvB=mv^2/r$. Show more…
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A deutron of kinetic energy $50 \mathrm{keV}$ is describing a circular orbit of radius $0.5 \mathrm{~m}$, is plane perpendicular to magnetic field $B$. The kinetic energy of proton that describes a circular orbit of radius $0.5 \mathrm{~m}$ in the same plane with the same magnetic field $B$, is (a) $200 \mathrm{keV}$ (b) $50 \mathrm{keV}$ (c) $100 \mathrm{keV}$ (d) $25 \mathrm{keV}$
Magnetic Effect of Current
Round 1
Knetic Energy An electron with kinetic energy $1.20 \mathrm{keV}$ circles in a plane perpendicular to a uniform magnetic field. The orbit radius is $25.0 \mathrm{~cm}$. Find (a) the speed of the electron, (b) the magnetic field, (c) the frequency, and (d) the period of the motion.
An electron of kinetic energy $1.20 \mathrm{keV}$ circles in a plane perpendicular to a uniform magnetic field. The orbit radius is $25.0 \mathrm{~cm}$. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.
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