Solution
Problem Analysis:
A ball (mass $m$, charge $q$) falls in a viscous fluid with terminal velocity $u$. A horizontal magnetic field $B$ is applied. The ball acquires a new terminal velocity $\vec{v}_t$. The power dissipated becomes $\eta = 0.5$ times the original power. We assume the viscous force acts opposite to velocity, but its functional dependence on speed is unknown.
Given Data: $m = 1.0\,\text{g} = 10^{-3}\,\text{kg}$, $q = 2.0\,\mu\text{C} = 2 \times 10^{-6}\,\text{C}$, $u = 0.1\,\text{m/s}$, $\eta = 0.5$.
Step 1: Force Equilibrium
At terminal velocity, the net force is zero. The three forces are: Gravity ($m\vec{g}$), Viscous force ($\vec{F}_v$), and Magnetic force ($\vec{F}_m = q\vec{v}_t \times \vec{B}$).
Since $\vec{F}_v$ is antiparallel to $\vec{v}_t$ and $\vec{F}_m$ is perpendicular to $\vec{v}_t$, these two forces are perpendicular to each other. Their vector sum must balance gravity.
$$ F_v^2 + F_m^2 = (mg)^2 $$Step 2: Power and Force Relations
Power dissipated is $P = \vec{F}_v \cdot \vec{v}_t = F_v v_t$.
Initially ($B=0$), $F_v = mg$, so $P_0 = mg u$.
With field $B$, $P = \eta P_0 = \eta mg u$. Thus, $F_v = \frac{P}{v_t} = \frac{\eta mg u}{v_t}$.
The magnetic force magnitude is $F_m = q v_t B$.
Step 3: Finding Maximum B
Substitute expressions for $F_v$ and $F_m$ into the equilibrium equation:
$$ \left( \frac{\eta mg u}{v_t} \right)^2 + (q v_t B)^2 = (mg)^2 $$Solving for $B^2$:
$$ q^2 B^2 v_t^2 = (mg)^2 – \frac{\eta^2 (mg)^2 u^2}{v_t^2} $$ $$ B^2 = \left(\frac{mg}{q}\right)^2 \left[ \frac{1}{v_t^2} – \frac{\eta^2 u^2}{v_t^4} \right] $$To find maximum $B$, we maximize the term in the brackets with respect to $v_t^2$. Let $x = 1/v_t^2$. We maximize $f(x) = x – \eta^2 u^2 x^2$.
The maximum occurs at $x = \frac{1}{2\eta^2 u^2}$. The maximum value is $\frac{1}{4\eta^2 u^2}$.
Thus,
$$ B_{max}^2 = \left(\frac{mg}{q}\right)^2 \frac{1}{4\eta^2 u^2} \implies B_{max} = \frac{mg}{2\eta q u} $$Given $\eta = 0.5$, we have $2\eta = 1$. So:
Calculation:
$$ B_{max} = \frac{10^{-3} \times 10}{2 \times 10^{-6} \times 0.1} = \frac{10^{-2}}{2 \times 10^{-7}} = 0.5 \times 10^5 = 50,000\,\text{T} $$