A small particle of mass m and charge Q is dropped in uniform horizontal magnetic field B . The maximum vertical displacement of particle is given by h = nm 2 g 2 Q 2 B 2 . Find the value of n .

Step-by-step solution

mgh = 1 2 mv 2 ⇒ v = 2 gh At lowest position, velocity becomes horizontal. ∴ v x = v = 2 gh Fsin ⁡ θ = ma x QvB sin ⁡ θ = ma x ⇒ QB vsin ⁡ θ = ma x ⇒ QB V y = ma x ⇒ QB dy dt = m dv x dt ⇒ QB ∫ 0 h dy = m ∫ 0 v x dv x ⇒ QBh = m v x ⇒ QBh = m 2 gh ⇒ h = 2 m 2 g Q 2 B 2