33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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46 NEET-AIPMT Chapterwise Topicwise Solutions Physics
qvB
mv 2
m × 2π
= ⇒ qB = mω
=
R
T
2πm
∴ T = is independent of v and R.
qB
16. (a) : When a charged particle having K.E., T enters
in a field of magnetic induction, which is perpendicular
to its velocity, it takes a circular trajectory. It does not
increase in energy, therefore T is the K.E.
17. (a) : If a moving charged particle is subjected to a
perpendicular uniform magnetic field, then according to
F = qvB sinq, it will experience a maximum force which
will provide the centripetal force to particle and it will
describe a circular path with uniform speed.
18. (c) : The frequency of revolution of a charged
particle in a perpendicular magnetic field is
υ = 1 = 1
π = v
π = v
2 2 2π × eB
T r v r
= eB
/
mv 2πm
2
mv
19. (d) : qvB sinθ=
R
5
mv
R = qB
= 3×
10
= . =
sinθ
10 × 03 . ×
1 002m 2 cm
8
2
20. (b) : When a positively charged particle enters in a
region of uniform magnetic field, directed perpendicular
to the velocity it experiences a centripetal force which
will move it in circular path with a uniform speed.
⎛ 1 2 ⎞
21. (a) : Kinetic energy of electron ×
⎝
⎜ mv
2 ⎠
⎟ = 10 eV
and magnetic induction (B) = 10 –4 Wb/m 2
1 −31 2 −19
Therefore, ( 91 . × 10 ) v = 10 × ( 16 . × 10 )
2
(. )
or, v 2 19
2 10 1 6 10
12
= × × × −
= 352 . × 10
−31
91 . × 10
or, v = 1.876 × 10 6 m/s
Centripetal force, mv 2
= Bev
r
−31 6
mv ( 91 . × 10 ) × ( 1. 876 × 10 )
Therefore, r = =
Be −4 −19
10 × (. 16×
10 )
=11 × 10 –2 m = 11 cm
mv
22. (c) : r = or r ∝v
qB
As v is doubled, the radius also becomes doubled. Hence
new radius = 2 × 2 = 4 cm
23. (d) : For a charged particle orbiting in a circular
path in a magnetic field mv 2
Bqr
= Bvq ⇒ v=
r
m
E mv Bqvr Bq r Bqr 2 2 2
Bqr
K = 1 2
= 1
2 2 = 2 ⋅ m
= 2m
2 2 2
Bq × r
Fordeuteron, E1
=
2×
2m
2 2 2
Bqr
For proton, E2
=
2m
E1
1 50 keV 1
= ⇒ = ⇒ E2
= 100 keV
E2 2 E2
2
24. (c) : Frequency, υ = eB m
or B = 2π υ
2πm
e
As mv 2
eBR
= evB or v = =2πυR
R
m
1 2 1
2
Kineticenergy, K = mv = m( 2πυ R)
= 2mp 2 u 2 R 2
2 2
25. (c)
26. (a) : Electron travelling in a magnetic field
perpendicular to its velocity follows a circular path.
27. (a) : In mass spectrometer when ions are accelerated
through potential V,
1 2
mv = qV
... (i)
2
where m is the mass of ion, q is the charge of the ion.
As the magnetic field curves the path of the ions in a
semicircular orbit
2
mv BqR
∴ Bqv = ⇒ v =
... (ii)
R
m
Substituting (ii) in (i), we get
2
1 ⎡
2
m BqR ⎤
q V
qV
2
⎢
m
⎥
⎣ ⎦
= or, m
= 2 2
BR
Since V, B are constants,
q
∝
1 charge on the ion
or,
∝
1
m 2 2
R mass of the ion
R
E
28. (a) : eE = evB ∴ v=
B
29. (d) : Electric field (E) = 20 V/m and magnetic field
(B) = 0.5 T.
The force on electron in a magnetic field = evB
Force on electron in an electric field = eE
Since the electron is moving with constant velocity,
therefore the resultant force on electron is zero.
i.e., eE = evB ⇒ v = E/B = 20/0.5 = 40 m s –1
30. (b) : According to Biot-Savart’s law,
⎛ dl r
× ⎞ µ 0 ⎛ dl × r ⎞
dB ∝ i ⎜ ⎟ or dB = i ⎜ ⎟.
⎝ 3
r ⎠ 4π
⎝ 3
r ⎠
31. (a) : Magnetic field due to i 1 = µ 01 i θ 1
2R 2π
(Into the plane)
Magnetic field due to
02 i 2
i2
= µ θ (Out of the plane)
2R
2π For parallel combination
Now, i 1 ρl2
A l2
= × =
i2
A ρl1
l1
1
i
R
⇒ 1 4 ( 2 π ) 1 i
= = ⇒
2
i1
= ⇒ i2 = 3i1
i 3
2 R
4 2 3 3
( π )