Advertisements
Advertisements
प्रश्न
A conducting bar is rotating with constant angular speed around a pivot at one end in a uniform magnetic field perpendicular to its plane of rotation. Obtain an expression for the rotational e.m.f. induced between the ends of the bar.
उत्तर
A conducting bar of length L rotates with constant angular speed ω about a pivot at one end in a uniform magnetic field B perpendicular to its plane of rotation.
The linear velocity at a distance r from the pivot is v = ωr
The small induced e.m.f. in an element dr is:
de = Bvdr = B(ωr)dr
To find the total e.m.f (e), integrate from r = 0 to r = L
`e = int_0^L Bω dr`
`e = Bω int_0^L r dr`
`e = Bω [r^2/2]_0^L`
`e = Bω L^2/2`
Expression for Induced e.m.f.:
`e = 1/2 BωL^2`
The induced rotational e.m.f. between the ends of the rotating conducting bar is `1/2BωL^2`.
The higher the angular speed, length of the bar, or magnetic field strength, the greater the induced e.m.f.
