Three long- straight and parallel wires are arranged as shown in figure- The force experienced by 10cm length of wire Q is1-4times -10-4-N toward the right1-4times -10-4-N toward the left2-6times -10-4-N toward the right2-6times -10-4-N toward the left

Question

Toppr Admin · Accepted Answer

Magnetic field produced by wire R at Q- B1-x3BC-0i12-x3C0-r1-x3BC-0-xD7-202-x3C0-xD7-2-xD7-10-x2212-2-

Three long, straight and parallel wires are arranged as shown in figure. The force experienced by $10 c m$ length of wire $Q$ is

$1.4 \times 10^{- 4} N$ toward the right
$1.4 \times 10^{- 4} N$ toward the left
$2.6 \times 10^{- 4} N$ toward the right
$2.6 \times 10^{- 4} N$ toward the left

Magnetic field produced by wire R at Q: $B_{1} = \frac{μ_{0} i_{1}}{2 π r_{1}} = \frac{μ_{0} \times 20}{2 π \times (2 \times 10^{- 2})}$

Three long, straight and parallel wires are arranged as shown in figure. The force experienced by 10cm length of wire Q is1.4×10−4N toward the right1.4×10−4N toward the left2.6×10−4N toward the right2.6×10−4N toward the left

Magnetic field produced by wire R at Q: B1=μ0i12πr1=μ0×202π×(2×10−2)

Three long, straight and parallel wires are arranged as shown in figure. The force experienced by $10 c m$ length of wire $Q$ is

$1.4 \times 10^{- 4} N$ toward the right
$1.4 \times 10^{- 4} N$ toward the left
$2.6 \times 10^{- 4} N$ toward the right
$2.6 \times 10^{- 4} N$ toward the left

Magnetic field produced by wire R at Q: $B_{1} = \frac{μ_{0} i_{1}}{2 π r_{1}} = \frac{μ_{0} \times 20}{2 π \times (2 \times 10^{- 2})}$