Question

10 A 10 A 2 B. The three identical wires of length 50 cm carry currents as shown in the Figure to right. Determine the magnitude and direction of the net force exerted on wire 1 by the other two wires. 2.0 cm 3 2.0 cm 10A 50 cm

Answer 1

The force between two long straight and parallel conductors separated by a distance r can be found by :

The force on a current-carrying wire in a magnetic field is F = IlB sin θ

To calculate B :

μ₀ = 4π × 10⁻⁷ H/m ≈ 12.57×10⁻⁷ H/m.

I = 10 A

r = 2×10⁻² m

B21 = (μ0 *I )/2*π*r

= (12.57×10⁻⁷  * 10 ) /( 2 * 3.14 * 2×10⁻² )

= 1.27 ×10⁻⁵    Tesla

sin θ = sin (180) = 0

F = IlB sin θ

= 10 * 0.5 *0

= 0

Similary in the next case too ; the angle is 0 so the force becomes zero .Since the velocity and magnetic field are parallel to each other, there is no orientation of your hand that will result in a force direction. Therefore, the force on this moving charge is zero.

P.S.If this helepd you please like the answer.Thankyou.