A square loop of side 20 cm and resistance \(1\,\Omega\) is moved towards right with a constant speed \(v_0\). The right arm of the loop is in a uniform magnetic field of 5 T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value \(4\,\Omega\). What should be the value of \(v_0\) so that a steady current of 2 mA flows in the loop?
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Only the right arm cuts the field, so emf \(= B\ell v_0\). Reduce the resistor network between the loop terminals P and Q to one equivalent resistance, add the loop resistance, and use \(\varepsilon = IR_{total}\).
Only the right arm (length \(\ell = 20\) cm \(= 0.2\) m) is inside the field, so the motional emf is \(\varepsilon = B\ell v_0 = 5 \times 0.2 \times v_0 = v_0\) volt. The loop is attached to the network at P and Q. On one side of P-Q the two \(4\,\Omega\) resistors are in series: \(4+4 = 8\,\Omega\). On the other side the other two \(4\,\Omega\) resistors are also in series: \(8\,\Omega\). These two \(8\,\Omega\) paths are in parallel: \(R_{net} = \frac{8\times 8}{8+8} = 4\,\Omega\). Total resistance in the circuit: \(R = 4 + 1 = 5\,\Omega\) (network plus loop). \(I = \frac{\varepsilon}{R} \Rightarrow 2\times 10^{-3} = \frac{v_0}{5} \Rightarrow v_0 = 10^{-2}\) m/s \(= 1\) cm/s. Answer: (2) 1 cm/s
