An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 105 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant. Electron v = 10 m/s P (1) 8 × 10-20 N (2) 4 × 10-20 N (3) 8x* 10-20 N (4) 4π * 10-20 N 20 cm 5 A [NEET-2021]
Answers & Comments
Answer:
The correct option is A
8
×
10
−
20
N
Magnetic field produced due to long current carrying wire at point
A
is
B
=
μ
0
4
π
2
I
r
B
=
10
−
7
×
2
×
5
20
×
10
−
2
=
1
2
×
10
−
5
T
⊙
Direction of magnetic field will be outward to the plane of paper.
Now, force acting on the electron due to this field,
→
F
=
q
(
→
v
×
→
B
)
|
→
F
|
=
1.6
×
10
−
19
×
10
5
×
1
2
×
10
−
5
=
0.8
×
10
−
19
∴
|
→
F
|
=
8
×
10
−
20
N
Hence,
(
A
)
is the correct option.
Answer:
The correct option is A
8
×
10
−
20
N
Magnetic field produced due to long current carrying wire at point
A
isB
=
μ
0
4
π
2
I
r
B
=
10
−
7
×
2
×
5
20
×
10
−
2
=
1
2
×
10
−
5
T
⊙
Direction of magnetic field will be outward to the plane of paper.
Now, force acting on the electron due to this field,
→
F
=
q
(
→
v
×
→
B
)
|
→
F
|
=
1.6
×
10
−
19
×
10
5
×
1
2
×
10
−
5
=
0.8
×
10
−
19
∴
|
→
F
|
=
8
×
10
−
20
N
Hence,
(
A
)
is the correct option.
Explanation: