Question

The magnetic field in a region is given by →B=→kB0Ly where L is a fixed length. A conducting rod of length L lies along the Y-axis between the origin and the point (0, L, 0). If the rod moves with a velocity v = v0 →i, find the emf induced between the ends of the rod.

Answer 1

Emf induced between the ends of the rod is$\frac{1}{2} B_{0} v_{0} L$

Explanation:

Step 1:

Magnetic field for the region in question,

$\vec{B}=\frac{B_{0}}{y} \hat{k}$

Step 2:

Rod length on a y-axis = L

rod velocity , $v=v_{0} i$

Step 3:

We'll find a tiny length dy feature on the rod.

In the element Emf is induced:

$d e=B v d y$

$d e=\frac{B_{0}}{L} y \times v_{0} \times d y$          

$=\frac{B_{0} v_{0}}{L} y d y$

$\text { And, } e=\frac{B_{0} v_{0}}{L} \int_{0}^{L} y d y$    

$=\frac{B_{0} v_{0}}{L}\left[\frac{y^{2}}{2}\right]_{0}^{L}$  

$=\frac{B_{0} v_{0}}{L} \frac{L^{2}}{2}=\frac{1}{2} B_{0} v_{0} L$