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

Correct answer: A. 20 ~A Force per unit length between two parallel current carrying wire is given by F= (μ_0)/(2 π) (I_1 I_2)/(r) Here, (μ_0)/(4 π)=10^ -7 NmA^ -2 , r=0.02 ~m F=2 × 10^ -3 ~N ~m^ -1 , I_1=10 ~A, I_2= ? So, 2 × 10^ -3 =2 × 10^ -7 × (10 × I_1)/(0.02) 0.02 × 10^4=10 I_1 I_1=20 ~A

AIIMS 2015

Exam: AIIMS
Year: 2015
Subject: Physics
Chapter: Magnetic Effects of Current

AIIMS 2015 · Physics · Magnetic Effects of Current

The force per unit length between two straight conductors separated...

The force per unit length between two straight conductors separated by a distance of

0.02 \mathrm{~m}

2 \times 10^{-3} \mathrm{~N} \mathrm{~m}^{-1}

. The current in one conductor is

10 \mathrm{~A}

and that in the other conductor is

Options

A $20 \mathrm{~A}$
B $5 \mathrm{~A}$
C $2 \mathrm{~A}$
D $10 \mathrm{~A}$

Correct answer

A. $20 \mathrm{~A}$

Step-by-step solution

Force per unit length between two parallel current carrying wire is given by F=

\frac{\mu_0}{2 \pi} \frac{I_1 I_2}{r}

Here,

\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{NmA}^{-2}, r=0.02 \mathrm{~m}

F=2

\times

10^{-3}

\mathrm{~N} \mathrm{~m}^{-1},

I_1=10

\mathrm{~A},

I_2=

\text{ ? }

So,

2 \times 10^{-3}=2 \times 10^{-7} \times \frac{10 \times I_1}{0.02}

0.02

\times

10^4=10 I_1 \quad \therefore I_1=20

\mathrm{~A}

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