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

The induced emf between two end is
`34.02 * 10^(-5)` V

Step-by-step explanation:

Given:

Length of rod
`l = 1` m

Height
`h = 1.23` m

Magnetic field
`B = 6.93 * 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 (m)/(s^(2) )`

`v = √(2gh)`

`v = √(2 * 9.8 * 1.23)`

`v = 4.91`
`(m)/(s)`

Put the velocity in above equation,

`\epsilon = 6.93 * 10^(-5) * 1 * 4.91`

`\epsilon = 34.02 * 10^(-5)` V

Therefore, the induced emf between two end is
`34.02 * 10^(-5)` V