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NEET 2021 Physics and Chemistry
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© examsnet.com
Question : 16
Total: 100
An infinitely long straight conductor carries a current of
5
A
as shown. An electron is moving with a speed of
10
5
m
∕
s
parallel to the conductor. The perpendicular distance between the electron and the conductor is
20
c
m
at an instant. Calculate the magnitude of the force experienced by the electron at that instant.
[NEET 2021]
4
×
10
−
20
N
8
π
×
10
−
20
N
4
π
×
10
−
20
N
8
×
10
−
20
N
Validate
Solution:
👈: Video Solution
Magnetic field produced due to current carrying wire at point 'A'
B
=
µ
0
4
π
2
l
r
B
=
10
−
7
×
2
×
5
20
×
10
−
2
=
1
2
×
10
−
5
(Tesla), upward to the plane of paper
Now, force acting on electron due to this field
→
F
=
q
(
→
v
×
→
B
)
|
→
F
|
=
1.6
×
10
−
19
×
10
5
×
1
2
×
10
−
5
=
0.8
×
10
−
19
N
|
→
F
|
=
8
×
10
−
20
N
© examsnet.com
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