Question 191: Magnetism and Matter

Question

An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field $$\overrightarrow B  = \left( {1.5 	imes {{10}^{ - 3}}T} ight)\widehat k$$
 at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is
detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on
the screen) is :
(electron’s charge = 1.6 × 10–19 C, mass of electron = 9.1 × 10–31 kg)

Accepted Answer

We know, R = $${{mv} \over {qB}}$$
= $${{\sqrt {2mK} } \over {qB}}$$
= $${{\sqrt {2 	imes 9.1 	imes {{10}^{ - 31}} 	imes 100 	imes 1.6 	imes {{10}^{ - 19}}} } \over {1.6 	imes {{10}^{ - 19}} 	imes 1.5 	imes {{10}^{ - 3}}}}$$
= $${\sqrt 5 }$$ cm
From diagram,
$$\sin 	heta  = {2 \over R}$$ = $${2 \over {\sqrt 5 }}$$
So $$	an 	heta  = {2 \over 1}$$
In $$\Delta $$MNQ,
$$	an 	heta  = {x \over 6}$$
$$ 	herefore $$ $${x \over 6}$$ = $${2 \over 1}$$
$$ \Rightarrow $$ $$x$$ = 12 cm
From diagram you can see,
QP = d > 12
By checking all the options you can see possible value of d = 12.87 cm

Answer

2.25 cm

Answer

12.87 cm

Answer

1.22 cm

Answer

11.65 cm

Magnetism and Matter

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