JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 4)
Photon with kinetic energy of 1MeV moves
from south to north. It gets an acceleration of
1012 m/s2 by an applied magnetic field (west to
east). The value of magnetic field : (Rest mass
of proton is 1.6 × 10–27 kg) :
0.71mT
7.1mT
0.071mT
71mT
Explanation
_8th_January_Morning_Slot_en_4_1.png)
K.E = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ 1 MeV = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ 1.6 $$ \times $$ 10-19 $$ \times $$ 106 = $${1 \over 2}$$ $$ \times $$ 1.6 × 10–27 $$ \times $$ v2
$$ \Rightarrow $$ v = $$\sqrt 2 \times {10^7}$$ m/s
Fm = qvB sin$$\theta $$
$$ \Rightarrow $$ Fm = qvB [as $$\theta $$ = 90o]
$$ \Rightarrow $$ ma = Bev
$$ \Rightarrow $$ 1.6 × 10–27 $$ \times $$ 1012 = 1.6 $$ \times $$ 10-19 $$ \times $$ $$\sqrt 2 \times {10^7}$$ $$ \times $$ B
$$ \Rightarrow $$ B = 0.71 mJ
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