Answer to Question #231454 in Electricity and Magnetism for Anuj

Question

A beam of protons, each of which has mass m, and charge qe moving with a velocity V enters perpendicularly a uniform magnetic field with the magnitude B= 2uT directed into plane of the page. After the beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction, the beam enters a uniform electric field with the intensity E = 2.10-8 V/m directed downward existing above a distance d=10 m. The beam travels a distance of 6 cm in the magnetic field. After the proton beam exits the magnetic field, what will be passing time till the proton beam stops in the electric field? (Neglect the effect of gravitation).

Answer 1

Since proton exists the magnetic field after travelling a quater circle, so the length of arc will traveled will be,

$L= \frac{2 \pi r}{4} \\ L= \frac{\pi r}{2}$

As per the question this length is equal to 6 cm.

$\frac{\pi r}{2}= 6 \times 10^{-2} \;m \\ \frac{3 r}{2} = 6 \times 10^{-2} \;m \\ r = 4 \times 10^{-2} \;m$

Radius of circle under went is equal to 4 cm. (π= 3)

Since Lorentz Magnetic force provides the necessary centripetal force,

$qvB = \frac{mv^2}{r} \\ r = \frac{mv}{qB}$

As seen from the figure the point of entry and exit for proton is equal to radius of circle under went.

$v=\frac{qBr}{m} \\ v = \frac{(1.6 \times 10^{-19})(2 \times 10^{-6}) (4 \times 10^{-2})}{1.67 \times 10^{-27}} \\ v = 7.66 \; m/s$

The proton enters and exits the region of magnetic field at 7.66 m/s as Lorentz force only produces change in direction.

The proton then moves 10 m in the region of no field, hence its kinetic energy hence velocity remains same, i.e proton enters electric field with a velocity of 7.66 m/s.

Once it enters the Electric field a Electric force, F = qE, opposes its motion hence,

$m_p\vec{a}= -q \vec{E}$

retardation produced,

$\vec{a} = -\frac{q \vec{E}}{m_p} \\ \vec{a} = - \frac{(1.6 \times 10^{-19})(2 \times 10^{-8})}{1.67 \times 10^{-27}} \\ \vec{a}= -1.92\; m/s$

Let S be the distance within the Electric field after which the proton comes to momentary rest.

By equation of motion,

$v^2=u^2 + 2aS \\ 0^2 = u^2+ 2aS \\ 0 = u^2 + 2(-1.92)S \\ u^2 = 3.84S \\ S=\frac{u^2}{3.84} \\ S = \frac {7.66^2}{3.92} \\ S = 14.97 \; m$

Time can be calculated using equation,

$v = u + at \\ 0=u+ at \\ t = \frac{ u}{a} \\ t = \frac{7.66}{3.92} \\ t = 1.95 \; s$

Answer: 1.95 s