Question

A small ball of mass m carrying positive charge +Q is dropped in uniform horizontal magnetic field B. The vertical depth of the deepest point of its path from initial position is $\frac{(3k-1)m^2g}{Q^2B^2}$. Find k. ('g' is acceleration due to gravity)

• A. 1
• B. 2
• C. 3
• D. None

Answer 1

Answer: (A)

1

Answer 2

The problem describes the motion of a charged ball dropped in a uniform horizontal magnetic field. We need to find the maximum vertical depth reached by the ball and use it to determine the value of 'k'.

1. Set up the Coordinate System and Forces:

Let the initial position of the ball be the origin (0,0,0).
Let the y-axis be vertically downwards.
Let the x-axis be horizontal, perpendicular to the magnetic field.
Let the z-axis be along the direction of the magnetic field B.
So, the magnetic field is $\vec{B} = B\hat{k}$.
The acceleration due to gravity is $\vec{g} = g\hat{j}$.

The forces acting on the ball are:

• Gravitational force: $\vec{F_g} = m\vec{g} = mg\hat{j}$
• Magnetic force: $\vec{F_m} = Q(\vec{v} \times \vec{B})$

Let the velocity of the ball be $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$.
The magnetic force will be:

$\vec{F_m} = Q( (v_x\hat{i} + v_y\hat{j} + v_z\hat{k}) \times B\hat{k} )$
$\vec{F_m} = Q(v_x B(\hat{i} \times \hat{k}) + v_y B(\hat{j} \times \hat{k}) + v_z B(\hat{k} \times \hat{k}))$
$\vec{F_m} = Q(v_x B(-\hat{j}) + v_y B(\hat{i}) + 0)$
$\vec{F_m} = QBv_y\hat{i} - QBv_x\hat{j}$

2. Equations of Motion:

Using Newton's second law, $m\vec{a} = \vec{F_{net}}$:

$m\frac{dv_x}{dt} = QBv_y \quad \cdots (1)$
$m\frac{dv_y}{dt} = mg - QBv_x \quad \cdots (2)$
$m\frac{dv_z}{dt} = 0 \quad \cdots (3)$

From equation (3), since the ball starts from rest ($v_z(0)=0$), $v_z$ remains zero throughout the motion. Thus, the motion is confined to the x-y plane.

3. Solve the Differential Equations:

Differentiate equation (1) with respect to time:

$m\frac{d^2v_x}{dt^2} = QB\frac{dv_y}{dt}$

From equation (2), $\frac{dv_y}{dt} = \frac{mg - QBv_x}{m}$.
Substitute this into the differentiated equation (1):

$m\frac{d^2v_x}{dt^2} = QB\left(\frac{mg - QBv_x}{m}\right)$
$\frac{d^2v_x}{dt^2} = \frac{QBg}{m} - \frac{Q^2B^2}{m^2}v_x$
$\frac{d^2v_x}{dt^2} + \left(\frac{QB}{m}\right)^2 v_x = \frac{QBg}{m}$

Let $\omega = \frac{QB}{m}$. The equation becomes:

$\frac{d^2v_x}{dt^2} + \omega^2 v_x = \omega g$

This is a second-order linear non-homogeneous differential equation.
The general solution is $v_x(t) = v_{xh}(t) + v_{xp}(t)$.
The homogeneous solution is $v_{xh}(t) = A\cos(\omega t) + C\sin(\omega t)$.
For the particular solution, assume $v_{xp} = K$ (constant). Then $\omega^2 K = \omega g \implies K = \frac{g}{\omega} = \frac{mg}{QB}$.
So, $v_x(t) = A\cos(\omega t) + C\sin(\omega t) + \frac{mg}{QB}$.

Now, find $v_y(t)$ using equation (1): $v_y = \frac{m}{QB}\frac{dv_x}{dt} = \frac{1}{\omega}\frac{dv_x}{dt}$.
$v_y(t) = \frac{1}{\omega}(-A\omega\sin(\omega t) + C\omega\cos(\omega t))$
$v_y(t) = -A\sin(\omega t) + C\cos(\omega t)$.

4. Apply Initial Conditions:

At $t=0$, the ball is dropped, so $v_x(0)=0$ and $v_y(0)=0$.
From $v_x(0)=0$: $A\cos(0) + C\sin(0) + \frac{mg}{QB} = 0 \implies A + \frac{mg}{QB} = 0 \implies A = -\frac{mg}{QB}$.
From $v_y(0)=0$: $-A\sin(0) + C\cos(0) = 0 \implies C = 0$.

Substitute A and C back into the velocity expressions:

$v_x(t) = -\frac{mg}{QB}\cos(\omega t) + \frac{mg}{QB} = \frac{mg}{QB}(1 - \cos(\omega t))$
$v_y(t) = -(-\frac{mg}{QB})\sin(\omega t) = \frac{mg}{QB}\sin(\omega t)$

5. Find the Maximum Vertical Depth ($y_{max}$):

The deepest point in the path is reached when the vertical velocity $v_y$ momentarily becomes zero.

$v_y(t) = \frac{mg}{QB}\sin(\omega t) = 0$

Since $\frac{mg}{QB} \neq 0$, we must have $\sin(\omega t) = 0$.
The first time this occurs after $t=0$ is when $\omega t = \pi$.
So, $t_{deepest} = \frac{\pi}{\omega} = \frac{\pi m}{QB}$.

Now, we can find the vertical position $y(t)$ by integrating $v_y(t)$:

$y(t) = \int_0^t v_y(t') dt' = \int_0^t \frac{mg}{QB}\sin(\omega t') dt'$
$y(t) = \frac{mg}{QB} \left[ -\frac{\cos(\omega t')}{\omega} \right]_0^t$
$y(t) = \frac{mg}{QB\omega} [-\cos(\omega t) - (-\cos(0))]$
$y(t) = \frac{mg}{QB\omega} [1 - \cos(\omega t)]$

Substitute $\omega = \frac{QB}{m}$:

$y(t) = \frac{mg}{QB(QB/m)} [1 - \cos(\omega t)]$
$y(t) = \frac{m^2g}{Q^2B^2} [1 - \cos(\omega t)]$

To find the maximum depth, substitute $t_{deepest} = \frac{\pi}{\omega}$:

$y_{max} = \frac{m^2g}{Q^2B^2} [1 - \cos(\pi)]$
$y_{max} = \frac{m^2g}{Q^2B^2} [1 - (-1)]$
$y_{max} = \frac{2m^2g}{Q^2B^2}$

6. Determine 'k':

The problem states that the vertical depth of the deepest point is $\frac{(3k-1)m^2g}{Q^2B^2}$.
Equating our derived $y_{max}$ with the given expression:

$\frac{(3k-1)m^2g}{Q^2B^2} = \frac{2m^2g}{Q^2B^2}$

$(3k-1) = 2$
$3k = 3$
$k = 1$