Solution to Question 7
(a) Stationary Shape
Consider a differential element of the wire subtending an angle $d\theta$. The magnetic force $dF_m$ acts radially outward (normal to the wire) due to the cross product $\vec{I} \times \vec{B}$.
Balancing the outward magnetic force with the inward component of tension $T$:
$$ dF_m = I (R d\theta) B $$ $$ 2 T \sin(d\theta/2) \approx T d\theta $$ $$ T d\theta = I R B d\theta \implies T = I R B $$Since $T$, $I$, and $B$ are constant throughout the wire, the radius of curvature $R$ must be constant. Thus, the wire assumes the shape of a circular arc.
(b) Range of Force Constant k
The tension is also provided by the elasticity of the wire (Hooke’s Law):
$$ T = k (L_{arc} – l_0) $$Substituting $T = IRB$:
$$ I R B = k (2\pi R – l_0) \quad \text{(considering the limiting case of a full circle)} $$For the wire to maintain a stable stationary shape between the nails without expanding indefinitely (or “breaking” the geometry), the restoring elastic force must be capable of balancing the magnetic expansion.
Rearranging for the equilibrium radius $R$:
$$ R (2\pi k – IB) = k l_0 $$ $$ R = \frac{k l_0}{2\pi k – IB} $$For a physical solution where the radius $R$ is positive and finite, the denominator must be positive:
$$ 2\pi k – IB > 0 $$