Solution
Consider a small element $d\vec{l}$ of the current-carrying wire. The magnetic force acting on this element is given by: $$ d\vec{F}_m = I (d\vec{l} \times \vec{B}) $$ Since the wire is flexible and carries a current $I$ in a uniform magnetic field $\vec{B}$, this magnetic force acts perpendicular to the length of the wire at every point. Under the combined influence of tension $T$ and the magnetic force $d\vec{F}_m$, the wire will curve until the components of tension balance the magnetic force.
For a wire with negligible mass (light wire) under constant tension $T$ in a uniform magnetic field, the radius of curvature $R$ is determined by the equilibrium condition normal to the wire: $$ T = B I R \implies R = \frac{T}{BI} $$ Since $T$, $B$, and $I$ are constant throughout the segment, the radius of curvature $R$ is constant. Therefore, each straight segment will assume the shape of a circular arc.
Let the wire form an arc subtending an angle at the center. We define $\theta$ as the angle the tangent to the wire makes with the vertical at the hooks.
1. Geometric Relations:
The loop consists of two segments, each of length $l$. The chord connecting the top and bottom hooks has length $y$. From the geometry of a circular arc: $$ \text{Arc Length } l = R(2\theta) \quad \text{(where } 2\theta \text{ is the total angle subtended at the center)} $$ $$ \text{Chord Length } y = 2R \sin \theta $$ Dividing these two gives the relation between the vertical separation $y$ and the angle $\theta$: $$ y = l \frac{\sin \theta}{\theta} $$
2. Force Equilibrium:
Consider the equilibrium of the block of mass $m$. The vertical components of the tension $T$ from both sides of the loop support the weight. The angle with the vertical at the connection point is $\theta$. $$ 2T \cos \theta = mg \implies T = \frac{mg}{2 \cos \theta} $$ We also know the relation between tension, magnetic field, and radius for a current-carrying wire: $$ T = B I R $$ Substituting $R = \frac{l}{2\theta}$: $$ T = \frac{BI l}{2\theta} $$
3. Maximizing the Lift:
Equating the expressions for Tension $T$:
$$ \frac{mg}{2 \cos \theta} = \frac{BI l}{2\theta} $$
$$ I = \frac{mg \theta}{B l \cos \theta} $$
The height to which the block is lifted is the difference between the initial length (when straight, $y=l$) and the final chord length $y$.
$$ \Delta h = l – y = l \left( 1 – \frac{\sin \theta}{\theta} \right) $$
To maximize the lift $\Delta h$, we must minimize the chord length $y$, which requires maximizing $\theta$.
From the current equation $I \propto \frac{\theta}{\cos \theta}$, as the current $I \rightarrow \infty$, $\cos \theta \rightarrow 0$, which implies $\theta \rightarrow \frac{\pi}{2}$.
At $\theta = \frac{\pi}{2}$, the wires form a semi-circle. The maximum lift is:
$$ \Delta h_{\text{max}} = l \left( 1 – \frac{\sin(\pi/2)}{\pi/2} \right) = l \left( 1 – \frac{1}{\pi/2} \right) $$
We are required to find the current $I$ that lifts the block to a height:
$$ h = l \left( 1 – \frac{3}{\pi} \right) $$
Comparing this to our general lift formula $\Delta h = l \left( 1 – \frac{\sin \theta}{\theta} \right)$, we have:
$$ \frac{\sin \theta}{\theta} = \frac{3}{\pi} $$
We test standard angles to solve this transcendental equation. Let’s try $\theta = \frac{\pi}{6}$ ($30^\circ$):
$$ \frac{\sin(\pi/6)}{\pi/6} = \frac{1/2}{\pi/6} = \frac{1}{2} \cdot \frac{6}{\pi} = \frac{3}{\pi} $$
This matches perfectly. So, the operating angle is $\theta = \frac{\pi}{6}$.
Now, substitute $\theta = \frac{\pi}{6}$ into the current equation derived in Part (b):
$$ I = \frac{mg \theta}{B l \cos \theta} $$
$$ I = \frac{mg (\pi/6)}{B l \cos(\pi/6)} = \frac{mg \pi}{6 B l (\frac{\sqrt{3}}{2})} $$
$$ I = \frac{mg \pi}{3\sqrt{3} B l} $$