Physics Circle home Student Login

Home›Chapters›Moving Charges and Magnetism PYQs›Force on Current-Carrying Conductors›Question 38

JEE Main 2022 (28 Jun, Shift 2) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2022 (28 Jun, Shift 2)

Q38JEE Main 2022 · 28 JUN · SHIFT 2Easy

Force on Current-Carrying Conductors

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 10−-6 N, then the value of x is approximately:

Show answer and solution

Answer: (c)

  1. Fl=μ0x22πd⇒2×10−6=2×10−7 x20.20\frac{F}{l} = \frac{\mu_0 x^2}{2\pi d} \Rightarrow 2\times10^{-6} = \frac{2\times10^{-7}\,x^2}{0.20}
  2. x2=2⇒x=2≈1.4x^2 = 2 \Rightarrow x = \sqrt{2} \approx 1.4
  3. So the answer is (c) 1.4.
Report a mistake

More questions on Force on Current-Carrying Conductors