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?

We need to determine the constant speed $$v_0$$ of a square loop so that a steady current of $$2\text{ mA}$$ flows through it as it moves through a magnetic field and interacts with an external resistor network.

1. Analyze the External Resistor Network

From the layout described on the page, the loop is connected to an infinite or balanced grid network of resistors where each resistor has a value of $$R_0 = 4\ \Omega$$.

• For the standard infinite grid of square resistors, the equivalent resistance ($$R_{\text{net}}$$) measured across two adjacent terminals where the loop connects simplifies cleanly to:

  $$R_{\text{net}} = R_0 = 4\ \Omega$$

• The square loop itself has its own internal resistance:

  $$R_{\text{loop}} = 1\ \Omega$$

• Therefore, the total equivalent resistance ($$R_{\text{total}}$$) of the entire circuit is the sum of the loop resistance and the network resistance:

  $$R_{\text{total}} = R_{\text{loop}} + R_{\text{net}} = 1\ \Omega + 4\ \Omega = 5\ \Omega$$

2. Calculate the Induced Electromotive Force (emf)

As the square loop moves to the right with speed $$v_0$$, its right arm (of length $$l = 20\text{ cm} = 0.2\text{ m}$$) cuts through a uniform magnetic field ($$B = 5\text{ T}$$) going into the page. This produces a motional electromotive force ($$e$$) given by:

$$e = B \cdot l \cdot v_0$$

Substitute the given numerical values into the expression:

$$e = 5 \times 0.2 \times v_0 = 1 \cdot v_0 = v_0$$

3. Relate emf to the Steady Current

According to Ohm's law, the steady current ($$I$$) flowing through the circuit is equal to the induced emf divided by the total resistance:

$$I = \frac{e}{R_{\text{total}}}$$

We are given that the required steady current is $$I = 2\text{ mA} = 2 \times 10^{-3}\text{ A}$$. Substitute this and our resistance values into the formula:

$$2 \times 10^{-3} = \frac{v_0}{5}$$

$$v_0 = 5 \times 2 \times 10^{-3} = 10 \times 10^{-3} = 0.01\text{ m s}^{-1}$$

Converting the velocity from meters per second ($$\text{m s}^{-1}$$) to centimeters per second ($$\text{cm s}^{-1}$$):

$$v_0 = 0.01 \times 100 = 1\text{ cm s}^{-1}$$

Conclusion

The speed $$v_0$$ of the loop should be $$1\text{ cm s}^{-1}$$, which matches Option C.