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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 111
Total: 160
A small block of mass 20 g and charge 4 mC is released on a long smooth inclined plane of inclination 45°. A uniform horizontal magnetic field of 1 T is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is
2 s
3 s
5 s
6 s
Validate
Solution:
The free body diagram of the system is shown below,
The magnetic force on the particle is,
F
=
B
q
v
The equilibrium equation for the particle on inclined plane is,
F
=
m
g
cos
θ
B
q
v
=
m
g
cos
θ
v
=
m
g
cos
θ
q
B
The time taken to reached at velocity v is,
v
=
u
+
a
t
m
g
cos
θ
q
B
=
0
+
(
g
sin
θ
)
t
t
=
m
cot
θ
q
B
Substituting the values, we get
t
=
0.02
cot
45
°
4
×
10
−
3
×
1
t
=
5
s
© examsnet.com
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