Question

An angle $AOC$ made of a conducting wire moves along its bisector through a magnetic field $B$ with speed $v$ as given in the figure. Find the emf induced between the two free ends if the magnetic field is perpendicular to the plane. $(OA = OC = l)$

• A. $Blvsin(\frac{\theta}{2})$
• B. $2Blvsin(\frac{\theta}{2})$
• C. $Blvcos(\frac{\theta}{2})$
• D. $2Blvcos(\frac{\theta}{2})$

Answer 1

Answer: (B)

$2Blvsin(\frac{\theta}{2})$

Answer 2

The induced electric field in the moving conductor is $\vec{E}_{ind} = \vec{v} \times \vec{B}$. Since $\vec{v}$ is in the plane and $\vec{B}$ is perpendicular to the plane, $\vec{E}_{ind}$ is in the plane and perpendicular to $\vec{v}$. Let the bisector be along the x-axis, so $\vec{v} = v\hat{i}$. Let $\vec{B} = B\hat{k}$. Then $\vec{E}_{ind} = v\hat{i} \times B\hat{k} = -vB\hat{j}$. The emf induced along OA is $emf_{OA} = \int_O^A \vec{E}_{ind} \cdot d\vec{r}$. The component of $d\vec{r}$ along the direction of $\vec{E}_{ind}$ (y-axis) is $dr \sin(\theta/2)$. So, $emf_{OA} = \int_0^l (-vB\hat{j}) \cdot (dr(\cos(\theta/2)\hat{i} + \sin(\theta/2)\hat{j})) = \int_0^l -vB\sin(\theta/2) dr = -Blv\sin(\theta/2)$. The emf induced along OC is $emf_{OC} = \int_0^l (-vB\hat{j}) \cdot (dr(\cos(-\theta/2)\hat{i} + \sin(-\theta/2)\hat{j})) = \int_0^l -vB(-\sin(\theta/2)) dr = Blv\sin(\theta/2)$. The potential difference between A and C is $V_A - V_C = emf_{OA} - emf_{OC} = -Blv\sin(\theta/2) - Blv\sin(\theta/2) = -2Blv\sin(\theta/2)$. The magnitude of the induced emf is $2Blv\sin(\theta/2)$.