Question

2. Three long, straight parallel wires, carrying current are arranged as shown in the figure. The force experienced by a 25 cm length of wire C is (a) 10-3 N (b) 2.5 × 10-3 N (c) zero (d) 1.5 × 3 N

Answer 1

$\mathfrak{\huge{\underline{\underline{\red{QuEsTiOn\:?}}}}}$

✳ Three long, straight parallel wires, carrying current are arranged as shown in the figure. The force experienced by a 25 cm length of wire C is

(a) 10-3 N

(b) 2.5 × 10-3 N

(c) zero

(d) 1.5 × 3 N

$\mathcal{\huge{\underline{\underline{\green{AnSwEr:-}}}}}$

Option.C )➡ Zero

$\mathcal{\huge{\underline{\underline{\orange{SoLuTiOn:-}}}}}$

▶ The Following solution refers to the attachment. ✔

_____________________________

Additional information :-

• The SI unit of electric current is Ampere.

• The SI unit of Force is Newton.

• Parallel lines are the lines which never intersect each other.

• An electric wire is a conditing wire to carry electricity from one place to another.

• Mostly wire are made up of copper or alumnium.

☜☞☜☞☜☞☜☞☜☞☜☞☜☞☜☞

Answer 2

Correct Question:-

Three long, straight parallel wires, carrying current are arranged as shown in the figure. The force experienced by a 25 cm length of wire C is (a) 10-3 N (b) 2.5 × 10-3 N (c) zero (d) 1.5 × 3 N

The image is given below.

Your Answer:-

Let the magnetic field be B

So,

B on C due to D =

$\tt \dfrac{u_\circ}{2\pi s}\times I \\\\ \tt Here\:\:we \:\: have \:\:I=30A \\\\ \tt = \dfrac{u_\circ}{2\pi\times 3\times 10^{-2}}\times 30$

And

B on C due to G =

$\tt \dfrac{u_\circ}{2\pi s}\times I \\\\ \tt Here\:\:we \:\: have \:\:I=20A \\\\ \tt = \dfrac{u_\circ}{2\pi\times 2\times 10^{-2}}\times 20$

Now,

Force on C due to D =

$\tt i_c\times (\overrightarrow{length}\times \overrightarrow{B}) \\\\ \tt =i_c\times lb \\\\ \tt = 10 \times 25 \times 10^{-2} \times \dfrac{u_\circ}{2\pi\times 3\times 10^{-2}}\times 30 \\\\ \tt = 10 \times 25 \times 10^{-2} \times \dfrac{u_\circ}{2\pi }\times 1000$

Force on C due to G =

$\tt i_c\times (\overrightarrow{length}\times \overrightarrow{B}) \\\\ \tt =i_c\times lb \\\\ \tt = 10 \times 25 \times 10^{-2} \times \dfrac{u_\circ}{2\pi\times 2\times 10^{-2}}\times 20 \\\\ \tt = 10 \times 25 \times 10^{-2} \times \dfrac{u_\circ}{2\pi }\times 1000$

Both forces are equal but they are in different directions so net force is Zero

Option C is correct.