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.

Question

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. (A) 4 × 10-20 N (B) 8 π × 10-20 N (C) 4 π × 10-20 N (D) 8 × 10-20 N

Tardigrade Admin · Accepted Answer

B=(μ0 ii/2 π r) F=e v B=(e v × μ0 i/2 π r) =((1.6 × 10-19) × 105 × 4 π × 10-7 × 5/2 π × 0.2) N =(1.6 × 2 × 5/2) × 10-20 N ⇒ F=8 × 10-20 N

$4 \times 1 0^{- 20} N$

$8 π \times 1 0^{- 20} N$

$4 π \times 1 0^{- 20} N$

$8 \times 1 0^{- 20} N$

$B = \frac{μ _{0} i ^{i}}{2 π r}$
$F = e v B = \frac{e v \times μ _{0} i}{2 π r}$
$= \frac{( 1.6 \times 1 0 ^{- 19} ) \times 1 0 ^{5} \times 4 π \times 1 0 ^{- 7} \times 5}{2 π \times 0.2} N$
$= \frac{1.6 \times 2 \times 5}{2} \times 1 0^{- 20} N$
$\Rightarrow F = 8 \times 1 0^{- 20} N$

4×10−20N

8π×10−20N

4π×10−20N

8×10−20N

B=2πrμ0​ii​ F=evB=2πrev×μ0​i​ =2π×0.2(1.6×10−19)×105×4π×10−7×5​N =21.6×2×5​×10−20N ⇒F=8×10−20N

$4 \times 1 0^{- 20} N$

$8 π \times 1 0^{- 20} N$

$4 π \times 1 0^{- 20} N$

$8 \times 1 0^{- 20} N$

$B = \frac{μ _{0} i ^{i}}{2 π r}$
$F = e v B = \frac{e v \times μ _{0} i}{2 π r}$
$= \frac{( 1.6 \times 1 0 ^{- 19} ) \times 1 0 ^{5} \times 4 π \times 1 0 ^{- 7} \times 5}{2 π \times 0.2} N$
$= \frac{1.6 \times 2 \times 5}{2} \times 1 0^{- 20} N$
$\Rightarrow F = 8 \times 1 0^{- 20} N$