The circuit in figure consists of wires at the top and bottom and identical springs as the left and right sides. The wire at the bottom has a mass of $$10$$ g and is $$5$$ cm long. The wire is hanging as shown in the figure. The springs stretch $$0.5$$ cm under the weight of the wire and the circuit has a total resistance of $$12\Omega$$. When the lower wire is subjected to a static magnetic field, the springs, stretch an additional $$0.3$$ cm. The magnetic field is

$$k_{\text{eq}} = 2k$$

Given: $$m = 10^{-2}\text{ kg}$$, $$x_1 = 5 \times 10^{-3}\text{ m}$$, $$x_2 = 3 \times 10^{-3}\text{ m}$$, $$V = 24\text{ V}$$, $$R = 12\ \Omega$$, $$l = 5 \times 10^{-2}\text{ m}$$

Using initial equilibrium:

$$mg = k_{\text{eq}}x_1 \implies 10^{-2} \times 10 = k_{\text{eq}} \times (5 \times 10^{-3}) \implies k_{\text{eq}} = 20\text{ N/m}$$

Using Ohm's law: $$I = \frac{V}{R} = \frac{24}{12} = 2\text{ A}$$

Using magnetic force for additional stretch: $$F_m = k_{\text{eq}}x_2 \implies I l B = k_{\text{eq}}x_2$$

$$2 \times (5 \times 10^{-2}) \times B = 20 \times (3 \times 10^{-3}) \implies 0.1B = 0.06 \implies B = 0.6\text{ T}$$

Using right-hand rule for downward force with leftward current:

$$\vec{F}_m = I(\vec{l} \times \vec{B}) \implies -\hat{j} = (-\hat{i}) \times \vec{B} \implies \vec{B} = B\hat{k}$$