This time I1 = 5.03, I2 = 1.44 A, and the two wires are separated by 9.70 cm. Now consider the charge q = 7.27 x 10^-6 C, located a distance of 4.27 cm to the right of wire I2, moving to the right at speed v = 24.6 m/s. What is the magnitude of the total magnetic force on this charge? 1.19E-09 N 9.52E-10 N 2.49E-09 N 1.99E-09 N The diagram displays a pair of vertical wires with currents flowing upwards. The wire on the left carries a current labeled $I_1$, and the wire on the right carries a current labeled $I_2$. To the right of the second wire, there is a small circle labeled $q$, representing a charge, with an arrow pointing horizontally to the right labeled $v$, which signifies the velocity of the charge. This setup is often used in physics to illustrate the interactions between magnetic fields produced by current-carrying wires and moving charges, demonstrating concepts like the magnetic force exerted on moving charges in the presence of magnetic fields.

Given, This time I1 = 5.03 A, I2 = 1.44 A, and the two wires are separated d = 9.70 cm = 0.097 m.…

Answered: This time I1 = 5.03, I2 = 1.44 A, and…

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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This time I1 = 5.03, I2 = 1.44 A, and the two wires are separated by 9.70 cm. Now consider the charge q = 7.27 x 10^-6 C, located a distance of 4.27 cm to the right of wire I2, moving to the right at speed v = 24.6 m/s. What is the magnitude of the total magnetic force on this charge?

		1.19E-09 N
		9.52E-10 N
		2.49E-09 N
		1.99E-09 N

The diagram displays a pair of vertical wires with currents flowing upwards. The wire on the left carries a current labeled $I_1$, and the wire on the right carries a current labeled $I_2$. To the right of the second wire, there is a small circle labeled $q$, representing a charge, with an arrow pointing horizontally to the right labeled $v$, which signifies the velocity of the charge.

This setup is often used in physics to illustrate the interactions between magnetic fields produced by current-carrying wires and moving charges, demonstrating concepts like the magnetic force exerted on moving charges in the presence of magnetic fields.

Expert Solution

Step 1

Given, This time I₁ = 5.03 A, I₂ = 1.44 A, and the two wires are separated d = 9.70 cm = 0.097 m.

Now consider the charge q = 7.27 x 10^-6 C, located at a distance of r = 4.27 cm = 0.0427 m to the right of wire I₂, moving to the right at speed v = 24.6 m/s. To find the magnitude of the total magnetic force B on this charge.

Force (F) per unit length (L) between parallel conductors is given by:

$\frac{F}{L} = \frac{μ_{0} I_{1} I_{2}}{2 πd}$

Substituting the known values:

$\begin{array}{rcl} \frac{F}{L} & = & \frac{(4 π \times 10^{- 7}) (5.03) (1.44)}{2 π \times (0.097)} \\ = & 149.3 \times 10^{- 7} \frac{N}{m} \end{array}$

We know that force due to any current-carrying wire:

$F = l l B \sin θ S i n c e, θ = 90 ° F = l l B \sin 90 ° F = I l B$

$F_{2} = - F_{1} From Newton' s third law$

Force on wire 2 by magnetic field of wire 1:

$\begin{array}{rcl} F_{2} & = & I_{2} l B_{1} \\ B_{1} & = & \frac{F_{2}}{l} I_{2} \\ = & (149.3 \times 10^{- 7}) (1.44) \\ = & 14.99 \times 10^{- 7} T \end{array}$

Force on wire 1 by magnetic field of wire 2:

$\begin{array}{rcl} F_{1} & = & I_{1} l B_{2} \\ B_{2} & = & \frac{F_{1}}{l} I_{1} \\ = & (149.3 \times 10^{- 7}) (5.03) \\ = & 750.9 \times 10^{- 7} T \end{array}$

Magnetic field on the charge:

$B = \frac{μ_{0}}{4 π} \frac{q v}{r^{2}} B = \frac{4 π \times 10^{- 7}}{4 π} \frac{(7.27 \times 10^{- 6}) (24.6)}{{(0.0427)}^{2}} = 9.8 \times 10^{- 9} T$

Total magnetic field of the charge is the sum of all three magnetic fields:

$\begin{array}{rcl} B_{t o t} & = & 1499 \times 10^{- 9} + 75090 \times 10^{- 9} + 9.8 \times 10^{- 9} T \\ = & 99 \times 10^{- 9} T \end{array}$

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