Question

4. A cycle wheel of radius 0.5 m is rotated
with constant angular velocity of 10 rad/s
in a region of magnetic field of 0.1 T which
is perpendicular to the plane of the wheel.
The EMF generated between its centre and
the rim is,
(a) 0.25 V
(b) 0.125 V
(c) 0.5 V
(d) zero
why they are using E=1/2 BWR^2​

Answer 1

Answer:

Option B: 0.125 V

Explanation:

Given:

• Radius of the wheel = 0.5 m
• Angular velocity = 10 rad/s
• Magnetic field = 0.1 T

To Find:

• The emf generated between its centre and  the rim

Solution:

Given that the wheel is rotating with a constant angular velocity in a plane which is normal to the magnetic field.

Let us consider a rod of length l from the centre of the wheel.

The emf produced across the two ends of the rod is given by,

$\displaystyle \sf \epsilon =\int\limits{v\:B} \, dx$

The magnetic field is uniform and constant throughout,

$\displaystyle \sf \epsilon =B\int\limits{v} \, dx$

$\displaystyle \sf \epsilon =B\int\limits{\omega \times r} \, dr$

where ω is the angular velocity and r is the radius.

Here angular velocity is constant,

Therefore we get,

$\displaystyle \sf \epsilon =B\: \omega \int\limits{ r} \, dr$

$\boxed{\sf \epsilon = \dfrac{1}{2} \times Br^2\times \omega}$

Now substitute the given data,

$\sf \epsilon =\dfrac{1}{2} \times 0.1\times 10\times (0.5)^2$

$\sf \epsilon =\dfrac{1}{2} \times 1\times 0.25$

$\sf \implies 0.125\:V$

Hence the emf produced id 0.125 V.

Therefore option B is correct.