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

Here is your answer:

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

Given:

Resistance of wire, R = 2.5 Ω

B = 2 × 10⁻⁵ weber/m²

g = 10.0 m/s²

To Find:

The current in circuit, I = ?

Solution:

As we know,

⇒  $e=BIV$

On putting the estimated values, we get

⇒    $=2\times 10^5\times 500\times \sqrt{2}\times 10\times 5$

⇒    $=0.1 \ V$

Now,

Current,

⇒   $I=\frac{e}{R}$

⇒      $=\frac{0.1}{2.5}$

⇒      $=0.04 \ amp$

So that the induced current will be "0.04 amp".