Magnetic Field due to a current carrying wire at 20cm in terms of current I is given by,
B=2πdμ0I
B=20×10−22×10−7×5
B=0.5×10−5T
Now,
Force on charge particle =qvB
Fm=eVB
Fm=1.6×10−19×105×0.5×10−5
=8×10−20N
NEET UG 2021 — Physics Electromagnetism
2021mcqhard
An infinitely long straight conductor carries a current of 5A as shown. An electron is moving with a speed of 105ms−1 parallel to the conductor. The perpendicular distance between the electron and the conductor is 20cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.
Official previous-year question
Held on 30 Apr 2021 · Verified 9 Jul 2026.
Options
- A
4×10−20N
- B
8π×10−20N
- C
4π×10−20N
- D
8×10−20N
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