Question

In the figure, there is a uniform conducting structure in which each small square has side a. The structure is kept in uniform magnetic field B :-

$G=E$ $H=F$ $B=D$

• A. The magnetic force on the structure is $2\sqrt{2}$ iBa
• B. The potential of point B = potential of point D.
• C. Potential of point O = potential of point B
• D. The magnetic force on the structure is $\sqrt{2}$ iBa

Answer 1

Answer: (A), (B), (C)

A, B, C are correct

Answer 2

The problem asks us to analyze a uniform conducting structure (a 2x2 grid of squares) placed in a uniform magnetic field $\vec{B}$ directed inwards. Current $i$ enters at point A and leaves at point C. We need to evaluate the given options regarding magnetic force and potentials.

1. Magnetic Force on the structure:

The magnetic force on a current-carrying conductor in a uniform magnetic field is given by $\vec{F} = i (\vec{L}_{eff} \times \vec{B})$, where $\vec{L}_{eff}$ is the effective displacement vector from the point of current entry to the point of current exit.

In this case, current enters at A and leaves at C.

Let's set up a coordinate system. Let point B be at the origin (0,0). Then the coordinates of the points are: A = (0, 2a) B = (0, 0) C = (2a, 0) D = (2a, 2a) E = (0, a) F = (2a, a) G = (a, 2a) H = (a, 0) O = (a, a)

The effective displacement vector from A to C is $\vec{L}_{AC} = (2a - 0)\hat{i} + (0 - 2a)\hat{j} = 2a\hat{i} - 2a\hat{j}$. The magnetic field $\vec{B}$ is uniform and directed inwards (perpendicular to the plane), so $\vec{B} = -B\hat{k}$.

The magnetic force $\vec{F}$ is: $\vec{F} = i (\vec{L}_{AC} \times \vec{B})$ $\vec{F} = i ((2a\hat{i} - 2a\hat{j}) \times (-B\hat{k}))$ $\vec{F} = i (-2aB (\hat{i} \times \hat{k}) + 2aB (\hat{j} \times \hat{k}))$ Using the cross product rules ($\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$): $\vec{F} = i (-2aB (-\hat{j}) + 2aB (\hat{i}))$ $\vec{F} = i (2aB\hat{j} + 2aB\hat{i})$ $\vec{F} = 2aiB (\hat{i} + \hat{j})$

The magnitude of the magnetic force is: $|\vec{F}| = |2aiB (\hat{i} + \hat{j})| = 2aiB \sqrt{1^2 + 1^2} = 2aiB \sqrt{2}$ So, the magnetic force on the structure is $2\sqrt{2} iBa$. This matches option (A). Option (D) is incorrect.

2. Potentials of points:

The structure is a resistive network. Let the resistance of each segment of length 'a' be R. The current enters at A and leaves at C. The entire grid is symmetric about the diagonal AC. This means that points that are images of each other across the diagonal AC should have the same potential. The equation of the line AC (passing through (0,2a) and (2a,0)) is $x+y=2a$. To check symmetry for points (x,y) and (x',y') across $x+y=2a$:

1. The midpoint $((x+x')/2, (y+y')/2)$ lies on the line: $(x+x')/2 + (y+y')/2 = 2a \Rightarrow x+x'+y+y' = 4a$.
2. The line connecting (x,y) and (x',y') is perpendicular to $x+y=2a$. The slope of $x+y=2a$ is -1, so the slope of the perpendicular line is 1: $(y'-y)/(x'-x) = 1 \Rightarrow y'-y = x'-x$.

Let's check the points from the options:

(B) Potential of point B = potential of point D.

Point B is (0,0). Let's find its symmetric point (x',y'):

1. $0+x'+0+y' = 4a \Rightarrow x'+y' = 4a$.
2. $y'-0 = x'-0 \Rightarrow y'=x'$.

Substituting $y'=x'$ into the first equation: $x'+x'=4a \Rightarrow 2x'=4a \Rightarrow x'=2a$. Then $y'=2a$. So, the symmetric point to B(0,0) is (2a,2a), which is point D. Therefore, due to symmetry, $V_B = V_D$. This option is correct.

(C) Potential of point O = potential of point B.

Point O is (a,a). Point B is (0,0). Point O lies on the diagonal AC ($a+a=2a$), meaning it is on the axis of symmetry for current flow from A to C. Point B (0,0) does not lie on the diagonal AC ($0+0 \neq 2a$). Since current flows from A to C, there will be a potential drop along the path. If $V_O = V_B$, it would imply no current flows in the segment BO.

Let's consider the current distribution. Due to the symmetry about AC, we found $V_E = V_G$ and $V_B = V_D$. Also, current $I_{AE} = I_{AG} = i/2$. Current $I_{EO} = I_{GO}$. Current $I_{EB} = I_{GD}$. At node O, $I_{EO} + I_{GO} = I_{OH} + I_{OF}$. By symmetry, $I_{OH} = I_{OF}$. So $2I_{EO} = 2I_{OH} \Rightarrow I_{EO} = I_{OH}$. Similarly, current $I_{OH}$ flows from O to H, and $I_{OF}$ flows from O to F. Consider the path $A \to E \to B \to C$ and $A \to G \to D \to C$. These paths are symmetric. Consider the path $A \to E \to O \to H \to C$ and $A \to G \to O \to F \to C$. These paths are symmetric.

Let $V_C = 0$. The resistance of the path $A \to E \to O \to H \to C$ is $4R$. The resistance of the path $A \to E \to B \to C$ is $3R$. The resistance of the path $A \to G \to D \to C$ is $3R$. The resistance of the path $A \to G \to O \to F \to C$ is $4R$.

Let's find the equivalent resistance of the network. This is a common problem. For a 2x2 grid, the equivalent resistance between opposite corners (A and C) is $R_{eq} = \frac{5}{3}R$. The total current is $i$. $V_A = i R_{eq} = \frac{5}{3} iR$.

Currents: $I_{AE} = I_{AG} = i/2$. (This is not correct, $V_E = V_G$ implies $I_{AE} = I_{AG}$ only if paths from E,G to C are symmetric, which they are not in general. $V_E=V_G$ is correct from symmetry across AC.) Let's re-evaluate current distribution. Current entering A splits into $I_{AE}$ and $I_{AG}$. $I_{AE} = (V_A - V_E)/R$ and $I_{AG} = (V_A - V_G)/R$. Since $V_E=V_G$, $I_{AE}=I_{AG}$. So $I_{AE} = I_{AG} = i/2$. This is correct.

Now, at E: $i/2 = I_{EO} + I_{EB}$. At G: $i/2 = I_{GO} + I_{GD}$. Since $V_E=V_G$ and $V_B=V_D$, and $V_O$ is common for $I_{EO}$ and $I_{GO}$, it follows $I_{EO}=I_{GO}$ and $I_{EB}=I_{GD}$. At O: $I_{EO} + I_{GO} = I_{OH} + I_{OF}$. So $2I_{EO} = I_{OH} + I_{OF}$. At B: $I_{EB} + I_{HB} = I_{BC}$. At D: $I_{GD} + I_{FD} = I_{DC}$. At H: $I_{OH} + I_{BH} = I_{HC}$. At F: $I_{OF} + I_{DF} = I_{FC}$. At C: $I_{BC} + I_{HC} + I_{DC} + I_{FC} = i$.

From symmetry, $I_{OH}=I_{OF}$ and $I_{BH}=I_{DF}$ and $I_{HC}=I_{FC}$. So $2I_{EO} = 2I_{OH} \Rightarrow I_{EO} = I_{OH}$. And $I_{EB} + I_{BH} = I_{BC}$. $I_{EO} = I_{OH}$. $I_{EB} = I_{GD}$. $I_{BH} = I_{DF}$. $I_{BC} = I_{DC}$. $I_{HC} = I_{FC}$.

Let's apply KVL to loops or potentials. Let $V_C=0$. $V_H = I_{HC} R$. $V_B = I_{BC} R$. $V_F = I_{FC} R$. $V_D = I_{DC} R$. Since $I_{BC}=I_{DC}$ and $I_{HC}=I_{FC}$, we have $V_B=V_D$ and $V_H=V_F$. This is consistent with symmetry.

Consider the loop O-H-C-F-O. $V_O - V_H = I_{OH} R$. $V_H - V_C = I_{HC} R$. $V_F - V_C = I_{FC} R$. $V_O - V_F = I_{OF} R$. Since $V_H=V_F$ and $I_{OH}=I_{OF}$, these are consistent.

Now, let's find the current values. This is a standard problem for a 2x2 grid. Currents in a 2x2 grid between opposite corners (A and C): $I_{AE} = i/2$ $I_{AG} = i/2$ $I_{EO} = i/6$ $I_{GO} = i/6$ $I_{EB} = i/3$ $I_{GD} = i/3$ $I_{OH} = i/6$ $I_{OF} = i/6$ $I_{HB} = i/6$ $I_{FD} = i/6$ $I_{BC} = i/2$ $I_{DC} = i/2$ $I_{HC} = i/3$ $I_{FC} = i/3$

Now, let's calculate potentials relative to $V_C = 0$. $V_B = I_{BC} R = (i/2)R$. $V_D = I_{DC} R = (i/2)R$. (Consistent, $V_B=V_D$) $V_H = I_{HC} R = (i/3)R$. $V_F = I_{FC} R = (i/3)R$. (Consistent, $V_H=V_F$)

$V_O = V_H + I_{OH}R = (i/3)R + (i/6)R = (2i/6)R + (i/6)R = (3i/6)R = (i/2)R$. So, $V_O = (i/2)R$. Comparing $V_O$ and $V_B$: $V_O = (i/2)R$ $V_B = (i/2)R$ Therefore, $V_O = V_B$. This option is correct.

Since options (A), (B), and (C) are all correct, this is a multiple-correct-options question.