Question 263: Magnetism and Matter

Question

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 	imes 10^5 \mathrm{~ms}^{-1}$. When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is $x 	imes 10^4 \mathrm{~N} / \mathrm{C}$. The value of $x$ is _________. Take the mass of the proton $=1.6 	imes 10^{-27} \mathrm{~kg}$.

Accepted Answer

Let's break down the problem step by step:
When the proton moves undeflected in crossed electric and magnetic fields, the electric and magnetic forces balance each other. That is,
$$qE = qvB,$$
which simplifies to
$$E = vB.$$
After the electric field is switched off, the proton moves in a circular path under the action of the magnetic force. The magnetic force provides the required centripetal force:
$$qvB = \frac{mv^2}{r}.$$
Solving for the magnetic field $$B$$, we get:
$$B = \frac{mv}{qr}.$$
Now substitute this expression for $$B$$ back into the equilibrium condition:
$$E = vB = v\left(\frac{mv}{qr}ight) = \frac{mv^2}{qr}.$$
Plug in the given values:
Proton mass, $$m = 1.6 	imes 10^{-27} \, 	ext{kg}$$
Speed, $$v = 2 	imes 10^5 \, 	ext{m/s}$$
Radius, $$r = 2 \, 	ext{cm} = 0.02 \, 	ext{m}$$
Proton charge, $$q = 1.6 	imes 10^{-19} \, 	ext{C}$$
Thus,
$$E = \frac{(1.6 	imes 10^{-27} \, 	ext{kg})(2 	imes 10^5 \, 	ext{m/s})^2}{(1.6 	imes 10^{-19} \, 	ext{C})(0.02 \, 	ext{m})}.$$
Calculate the numerator:
$$(2 	imes 10^5)^2 = 4 	imes 10^{10},$$
So, $$(1.6 	imes 10^{-27}) 	imes (4 	imes 10^{10}) = 6.4 	imes 10^{-17}.$$
Calculate the denominator:
$$(1.6 	imes 10^{-19}) 	imes (0.02) = 3.2 	imes 10^{-21}.$$
Now compute the electric field:
$$E = \frac{6.4 	imes 10^{-17}}{3.2 	imes 10^{-21}} = 2 	imes 10^4 \, 	ext{N/C}.$$
The problem states that the magnitude of the electric field is $$x 	imes 10^4 \, 	ext{N/C}.$$ Since we found
$$E = 2 	imes 10^4 \, 	ext{N/C},$$
it follows that
$$x = 2.$$

Magnetism and Matter

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