Question

A horizontal telegraph wire 0.5 km long running east and west
in a part of a circuit whose resistance is 2.5 Ω. The wire falls to
g = 10.0 m/s² and B = 2 × 10⁻⁵ weber/m², then the current
induced in the circuit is
(a) 0.7 amp (b) 0.04 amp
(c) 0.02 amp (d) 0.01 amp

Answer 1

0.04 amp

Explanation:

We know that

e = BlV

 = 2 x 10^5 x 500 x √ 2x 10 x 5

 = 0.1 V

 As

I = e / R

= 0.1 / 2.5

=  0.04

Answer 2

Option(b) is the correct answer to the given question .

Explanation:

Given :

B = 2 × 10⁻⁵

I=0.5 km=500 M

Now we have to find the E which can be determined by the following formula.

$E = Blv$

=$2 * 10^{-5} * 500 * \sqrt{2} * 10 * 5$  

This will give the

E = 0.1 V

Now using the formula

$I =\frac{E}{R}$

$= \frac{0.1}{2.5}$

=  0.04 amp

Learn More :

• brainly.com/question/631125