Question

As shown in the figure, a rectangular loop of length $a$ and width $b$ and made of a conducting material of uniform cross-section is kept in a horizontal plane where a uniform magnetic field of intensity $B$ is acting vertically downward. The resistance per unit length of the loop is $\lambda \, \Omega / m$. If the loop is pulled with uniform velocity $v$ in a horizontal direction, which of the following statement(s) is/are true?

• A. Current in the loop I= Bbw / (2b=2a)
• B. Current will be in clockwise direction, looking from the top.
• C. Vp - Vs = VQ - V R' where V is the potential
• D. There cannot be any induction in part SR.

Answer 1

The Correct Option is A, B

We need to analyze the induced electromotive force (EMF), the induced current, the direction of the current, and the potential differences in the loop.

Step 1: Calculate the Induced EMF

As the loop is pulled through the magnetic field, an EMF is induced in the wires perpendicular to the velocity and magnetic field. The motional EMF induced in a wire of length \(L\) moving with velocity \(v\) perpendicular to a magnetic field \(B\) is given by: \[\mathcal{E} = BLv\]

In this case, the wires \(PQ\) and \(RS\) are the ones experiencing the induced EMF. Both wires have length \(b\), so the total induced EMF is: \[\mathcal{E} = Bvb\]

Step 2: Calculate the Resistance of the Loop

The loop has a total length of \(2(a + b)\). The resistance per unit length is \(\lambda\). Therefore, the total resistance \(R\) of the loop is: \[R = \lambda (2a + 2b) = 2\lambda (a + b)\]

Step 3: Calculate the Induced Current

Using Ohm's Law, the induced current \(I\) in the loop is: \[I = \frac{\mathcal{E}}{R} = \frac{Bvb}{2\lambda(a + b)}\]

Step 4: Determine the Direction of the Current

Using the right-hand rule (or Lenz's Law), as the loop moves to the right, the magnetic flux through the loop decreases (since the area within the field decreases). To oppose this change in flux, the induced current will create a magnetic field pointing downwards (in the same direction as the external field). This requires a clockwise current when viewed from above.

Step 5: Analyze the Potential Differences

Consider the potential difference \(V_P - V_S\) and \(V_Q - V_R\). Since PQ and RS are the source of the emf then \(V_P - V_Q=V_S-V_R=Bvb\) however due to the resistance, some of the potential will be lost hence \(V_P - V_S=V_Q-V_R\) is wrong.

There is definitely induction in part SR, since it is also moving through the magnetic field.

Step 6: Determine the Correct Statements

1. Current in the loop I= Bbv / (2b+2a): This statement is TRUE. Assuming the typo is \( \lambda \) instead of the number 1, and that the total length of the wire is \( 2(a+b) \) and thus total resistance is \( 2\lambda(a+b) \). The formula matches what we derived: \(I = \frac{Bvb}{2\lambda(a + b)}\)
2. Current will be in clockwise direction, looking from the top: This statement is TRUE.
3. V_P - V_S = V_Q - V_R, where V is the potential: This statement is FALSE.
4. There cannot be any induction in part SR: This statement is FALSE.

Conclusion

The correct statements are:

• Current in the loop I= Bbv / (2b+2a)
• Current will be in clockwise direction, looking from the top.