Question

Four pieces each of length $l$ of a conducting rod are joined as shown in the figure. The rod is placed in a downward magnetic field B. If the rod is moved towards right with a uniform velocity v, then the induced emf across the two ends of the rod is

• A. Blv
• B. 4 Blv
• C. 2 Blv sin $\left(\theta /2\right)$
• D. 2 Blv $(1+sin\theta /2)$

Answer 1

Answer: (D)

2 Blv $(1+sin\theta /2)$

From figure $b=lsin\left(\theta /2\right)$

Length of rod $L=l+2b+l=2l+2lsin\left(\theta /2\right)=2l[1+sin(\theta /2\left)\right]$

Induced emf between $A$ and $B$ $E=BvL$

$⟹$ $E=2lBv[1+sin(\theta /2\left)\right]$