JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Magnetic Effects of Current

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Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is

2 \times 1 0^{- 6} N

, then the value of x is approximately:

(A)1
(B)2.4
(C)1.4✓ Correct
(D)2

Answer: (C)

Solution

The force per unit length between two parallel current-carrying wires is given by

F / L = \frac{μ _{0} i _{1} i _{2}}{2 π d}

. Since both wires carry current

x

, we have:

\frac{F}{L} = \frac{μ _{0} x ^{2}}{2 π d}

Substituting the given values:

2 \times 1 0^{- 6} = \frac{4 π \times 1 0 ^{- 7} \times x ^{2}}{2 π \times 0.2}

Simplifying:

2 \times 1 0^{- 6} = \frac{2 \times 1 0 ^{- 7} \times x ^{2}}{0.2}

1 0^{- 6} = \frac{1 0 ^{- 7} \times x ^{2}}{0.2}

x^{2} = 10 \times 0.2 = 2

x = 2 \approx 1.4 A

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Magnetic Effects of Current

Solution

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