A short magnet oscillates with a time period 0.1 s at a place where horizontal magnetic field is 24 ~T . A downward current of 18 A is established in a vertical wire kept at a distance of 20 cm east of the magnet. The new time period of oscillations of the magnet is

Step-by-step solution

T ₁=0.1 ~s , ~B _ H =24 ~T =24 10⁻⁶ ~T , I =18 ~A , r =20 ~cm =20 10⁻² ~m Magnetic field due to vertical wire, B = ₀ I 2 r = 4 10⁻⁷ 18 2 20 10⁻² =18 ~T The time period of oscillation of magnet is aligned & T =2 I MB ~T 1 ~B & ~T ₂ ~T ₁ = B _ H ~B + B _ H = 24 18+24 = 2 7 & ~T ₂= 2 7 ~T ₁= 2 7 0.1=0.076 ~s aligned