A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :

### Related Formula vecF_B = q(vecv times vecB)$$\vec{F}_{B} = q(\vec{v} \times \vec{B})$$ vecF_net = mveca$$\vec{F}_{net} = m\vec{a}$$ ### Core Logic Since the solenoid is placed vertically, the magnetic field vecB$\vec{B}$ inside the solenoid will be parallel or anti-parallel to the vertical axis (either +y$+y$ or -y$-y$ axis). When the charged particle is released from rest, gravity pulls it vertically downward, meaning it gains velocity vecv$\vec{v}$ strictly along the y$y$-axis (parallel or anti-parallel to vecB$\vec{B}$). Because velocity and magnetic field are collinear (vecv parallel vecB$\vec{v} \parallel \vec{B}$ or vecv parallel -vecB$\vec{v} \parallel -\vec{B}$), the cross product vecv times vecB = 0$\vec{v} \times \vec{B} = 0$. Therefore, the magnetic force vecF_B = 0$\vec{F}_B = 0$. ### Step 1: Calculating Net Acceleration The only force acting on the particle is gravity. vecF_net = mvecg$$\vec{F}_{net} = m\vec{g}$$ a_net = g$a_{net} = g$

### Pattern Recognition Whenever a charged particle moves parallel to a magnetic field lines (like moving along the axis of a solenoid), the magnetic force is absolutely zero. It behaves like free fall. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism