Force per unit length between long parallel wires in the circuit is 3.6 × 10-3 N m-1. Resistance of the circuit is

Question

Force per unit length between long parallel wires in the circuit is 3.6 × 10-3 N m-1. Resistance of the circuit is (A) 3 Ω  (B) 1.5 Ω  (C) 4.5 Ω  (D) 6 Ω

Tardigrade Admin · Accepted Answer

Force per unit length between the parallel wires is given by,

F = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{d}

Here,

F = 3.6 \times 1 0^{- 3} N m^{- 1}, I_{1} = I_{2} = I

d = 8 mm = 8 \times 1 0^{- 3} m

So,

3.6 \times 1 0^{- 3} = 2 \times 1 0^{- 7} \times \frac{I ^{2}}{8 \times 1 0 ^{- 3}}

or,

I^{2} = 40 \times 3.6

or,

I = 12 A

Resistance of the circuit,

R = \frac{V}{I} = \frac{18}{12} = 1.5 O m e g a

Q. Force per unit length between long parallel wires in the circuit is $3.6 \times 1 0^{- 3} N m^{- 1}$ . Resistance of the circuit is

$3 Ω$

$1.5 Ω$

$4.5 Ω$

$6 Ω$

Q. Force per unit length between long parallel wires in the circuit is 3.6×10−3Nm−1. Resistance of the circuit is

3Ω

1.5Ω

4.5Ω

6Ω

Force per unit length between the parallel wires is given by, F=2πμ0​​dI1​I2​​ Here, F=3.6×10−3Nm−1,I1​=I2​=I d=8mm=8×10−3m So, 3.6×10−3=2×10−7×8×10−3I2​ or, I2=40×3.6 or, I=12A Resistance of the circuit, R=IV​=1218​=1.5Omega

Q. Force per unit length between long parallel wires in the circuit is $3.6 \times 1 0^{- 3} N m^{- 1}$ . Resistance of the circuit is

$3 Ω$

$1.5 Ω$

$4.5 Ω$

$6 Ω$