A metallic rod of length $l$  is tied to a string of length $2l$  and made to rotate with angular speed $\mathrm{\omega }$  on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f induced across the ends of the rod is 

In this problem, we are given a metallic rod of length l that is tied to a string of length 2l and rotates with angular speed ω on a horizontal table. One end of the string is fixed, and a vertical magnetic field B is applied in the region. The task is to find the induced electromotive force (e.m.f) across the ends of the metallic rod of length l.

Step-by-Step Analysis

To analyze the situation, let us consider a small element of the metallic rod of length l at a distance x from the fixed end of the string. The length of this small element is denoted by dx.

The velocity of this small element can be expressed as:

    v = ωx
  

Now, the induced e.m.f in this element of the rod can be calculated using the formula for motional e.m.f, which is the product of the magnetic field, the velocity, and the length of the element. The e.m.f induced in the element is:

    dε = B (ωx) dx
  

Where:

• B is the magnetic field strength,
• ω is the angular speed of the metallic rod of length l,
• x is the distance from the fixed end of the metallic rod of length l, and
• dx is the infinitesimal length of the rod element.

To find the total e.m.f induced across the ends of the metallic rod of length l, we integrate this expression from x = 2l (the fixed end) to x = 3l (the other end of the rod), which are the limits of the metallic rod of length l attached to the string. The total e.m.f, ε, is given by the integral:

    ε = ∫2l3l Bωx dx
  

Performing the integration:

    ε = Bω ∫2l3l x dx
  

Solving the integral:

    ε = Bω [ (1/2) x² ]2l3l
  

Now, we substitute the limits of integration:

    ε = Bω [ (1/2) (3l)² - (1/2) (2l)² ]
  

Simplifying the expression:

    ε = Bω [ (1/2) (9l²) - (1/2) (4l²) ]
  

Which simplifies further to:

    ε = Bω (5l²/2)
  

Final Result

Thus, the total e.m.f induced across the ends of the metallic rod of length l is:

    ε = (5/2) Bω l²