AP EAMCET20228 Jul 2022Evening ShiftPhysicsMagnetic Effects of CurrentActual
A long wire lies along X -axis and carries a current of 40 ~A in positive x -direction. A second long wire is perpendicular to the x y -plane, passes through point (3.0 ~m ) j and carries a current along positive z -direction. If the magnitude of resultant magnetic field at the point (2.0 ~m ) j R=5 10⁻⁶ ~T then the current in the second wire is (Permeability of free space, ₀=4 10⁻⁷ SI unit)
Options
- A30 ~A
- B15 ~A
- C25 ~A
- D7.5 ~A
Correct answer
B. 15 ~A
Step-by-step solution
Given arrangement is There are two magnetic fields at point 2 j as shown B₁= magnetic field due to wire A . = ₀ I₁ 2 d = ₀ 40 2 2 ~T and B₂= magnetic field due to wire B= ₀ I₂ 2 d = ₀ I₂ 2 1 ~T Resultant field magnitude, aligned R & = B₁^2+B₂^2 & = ₀ 2 20^2+I₂ (Given, R=5 10⁻⁶ ~T ) aligned aligned & 20^2+I₂^2 = 5 10⁻⁶ 2 10⁻⁷ & 400+I₂^2=625 aligned I₂^2=225 I₂=15 ~A