The force per unit length between two straight conductors separated by a distance of 0.02 m is 2 × 1 0 − 3 N m − 1 . The current in one conductor is 10 A and that in the other conductor is

Force per unit length between two parallel current carrying wire is given by F = 2 π μ 0 r I 1 I 2 Here, 4 π μ 0 = 1 0 − 7 NmA − 2 , r = 0.02 m F = 2 × 1 0 − 3 N m − 1 , I 1 = 10 A , I 2 = ? So, 2 × 1 0 − 3 = 2 × 1 0 − 7 × 0.02 10 × I 1 0.02 × 1 0 4 = 10 I 1 ∴ I 1 = 20 A

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PhysicsMagnetismMagnetic torque & energyMedium2 minPYQ_2015

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The force per unit length between two straight conductors separated by a distance of $0.02 m$ is $2 \times 1 0^{- 3} N m^{- 1}$ . The current in one conductor is $10 A$ and that in the other conductor is

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Force per unit length between two parallel current carrying wire is given by $F = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}$ Here, $\frac{μ _{0}}{4 π} = 1 0^{- 7} NmA^{- 2}, r = 0.02 m$ $F = 2 \times 1 0^{- 3} N m^{- 1}, I_{1} = 10 A, I_{2} = ?$ So, $2 \times 1 0^{- 3} = 2 \times 1 0^{- 7} \times \frac{10 \times I _{1}}{0.02}$ $0.02 \times 1 0^{4} = 10 I_{1} ∴ I_{1} = 20 A$

Stream:AIIMSSubject:PhysicsTopic:MagnetismSubtopic:Magnetic torque & energy

⏱ 2mℹ️ Source: PYQ_2015

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