Question

A 1.0 meter long rod is held horizontally in the east-west direction on a distant planet and is dropped from a height of 1.23 m. The horizontal component of the planet's magnetic field is 6.93e-05 T. Find the emf induced between the ends of the rod just before it hits the ground.

Answer 1

Answer:

The induced emf between two end is $34.02 \times 10^{-5}$ V

Explanation:

Given:

Length of rod $l = 1$ m

Height $h = 1.23$ m

Magnetic field $B = 6.93 \times 10^{-5}$ T

For finding induced emf,

  $\epsilon = Blv$

Where $v =$ velocity of rod,

For finding the velocity of rod.

From kinematics equation,

 $v^{2} = v_{o} ^{2} + 2gh$

Where $v_{o} =$ initial velocity, $g = 9.8 \frac{m}{s^{2} }$

   $v = \sqrt{2gh}$

   $v = \sqrt{2 \times 9.8 \times 1.23}$

   $v = 4.91$ $\frac{m}{s}$

Put the velocity in above equation,

   $\epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91$

   $\epsilon = 34.02 \times 10^{-5}$ V

Therefore, the induced emf between two end is $34.02 \times 10^{-5}$ V