Question

Four perpendicular long conducting wires are moving with speeds $v$, $2v$, $3v$ and $4v$ as shown in figure. If resistance per unit length of wires is $\lambda \frac{\Omega}{m}$ then induced emf in loop at $t=\frac{a}{v}$ is $NBav$, then $N$ is (given at $t=0$, loop of wires was a square having side of a)

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

The induced emf in a moving conductor is given by $\mathcal{E} = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$. Assuming a uniform magnetic field $\vec{B}$ into the page. Let the velocities be: Top wire: $v_1 = v\hat{j}$ Right wire: $v_2 = 2v\hat{i}$ Bottom wire: $v_3 = -3v\hat{j}$ Left wire: $v_4 = -4v\hat{i}$

At time $t$, the width $W(t) = a + (2v - (-4v))t = a + 6vt$. The height $H(t) = a + (v - (-3v))t = a + 4vt$.

The induced emfs in each segment (considering clockwise traversal): Top wire: $\mathcal{E}_1 = vB(a+6vt)$ Right wire: $\mathcal{E}_2 = -2vB(a+4vt)$ Bottom wire: $\mathcal{E}_3 = 3vB(a+6vt)$ Left wire: $\mathcal{E}_4 = -4vB(a+4vt)$

Total emf $\mathcal{E}_{total} = \mathcal{E}_1 + \mathcal{E}_2 + \mathcal{E}_3 + \mathcal{E}_4$ $\mathcal{E}_{total} = vB(a+6vt) - 2vB(a+4vt) + 3vB(a+6vt) - 4vB(a+4vt)$ $\mathcal{E}_{total} = vB [ (a+6vt) - 2(a+4vt) + 3(a+6vt) - 4(a+4vt) ]$ $\mathcal{E}_{total} = vB [ a+6vt - 2a-8vt + 3a+18vt - 4a-16vt ]$ $\mathcal{E}_{total} = vB [ (1-2+3-4)a + (6-8+18-16)vt ]$ $\mathcal{E}_{total} = vB [ -2a + 0vt ] = -2vBa$.

The magnitude is $|-2vBa| = 2vBa$. Given induced emf is $NBav$. Comparing, $N=2$.