Magnetic Field Due to a Current in a Long Straight Wire Problems and Solutions 2

 Problem#1

In Fig. 1, two long straight wires at separation d = 16.0 cm carry currents i₁ = 3.61 mA and i₂ 3.00i₁ out of the page. (a) Where on the x axis is the net magnetic field equal to zero? (b) If the two currents are doubled, is the zero-field point shifted toward wire 1, shifted toward wire 2, or unchanged?

Fig.1

Answer:
B net can only equal zero between the two wires, B₁ point upwards and B₂ point downward.
B_net = B₁ – B₂ = 0
Or, B₁ = B₂
μ₀i₁/2πr₁ = μ₀i₂/2πr₂
i₁/r = 3.00i₁/(d – r)
3r = d – r
r = d/4 = 16.0 cm/4 = 4.00 cm
so, when B = 0 at r = 4.00 cm

Fig.2

(b) If the two currents are doubled, is the zero-field point shifted toward wire 1, shifted toward wire 2, or unchanged?

As long as i₂ = 3.00i₁ the location of B­_net = 0 doesn’t change.

Problem#2
In Fig. 3, point P₂ is at distance R = 13.1 cm on the perpendicular bisector of a straight wire of length L = 18.0 cm carrying current i = 58.2 mA. (Note that the wire is not long.) What is the magnitude of the magnetic field at P₁ due to i?

Fig.3

Answer:
Consider the Biot-Savart law in differential form:

dB = (μ₀ ids x r)/(4πr³), with dB from all segments of wire point out of page.
dB = (μ₀ ids sin θ)/(4πr²)
B = (μ₀i/4π)∫[ds sin θ/r²]

Fig.4

With ds → dx and r² = R² + x², then sin θ = R/(R² + x²)^1/2

Known:
L = 0.18 m, R = 0.131 m and i = 0.0582 A, then

B = 5.03 x 10^-8 T = 50.3 nT, out of page

Problem#3
In Fig. 5, point P₂ is at distance R = 25.1 cm from one end of a straight wire of length L = 13.6 cm carrying current i = 0.693 A. (Note that the wire is not long.) What is the magnitude of the magnetic field at P₂?

Answer:
Consider the Biot-Savart law in differential form:

dB = (μ₀ ids x r)/(4πr³), with dB from all segments of wire point out of page.
dB = (μ₀ ids sin θ)/(4πr²)
B = (μ₀i/4π)∫[ds sin θ/r²]

With ds → dx and r² = R² + x², then sin θ = R/(R² + x²)^1/2

Known:
L = 0.18 m, R = 0.131 m and i = 0.0582 A, then

B = 1.32 x 10^-8 T = 132 nT, out of page

Problem#4
Figure 5 shows two very long straight wires (in cross section) that each carry a current of 4.00 A directly out of the page. Distance d₁ = 6.00 m and distance d₂ = 4.00 m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?

Fig.5

Answer:
The magnitude of the magnetic field at point P due to two wires is

B_P = B₁ sin α + B₂ sin α

Fig.6

with the magnitude of the magnetic field each wire at point P is

B₁ = B₂ = μ₀i/2πr

Then

B_P = (μ₀i/πr)sin α

from the figure we know

tan θ = ¾ or θ = 36.9⁰, so that α = 90 ─ 36.9 = 53.1⁰

so

B_P = [(4π x 10^-7 Wb/A.m)(4.00 A)/(2π x 5.00 m)] sin 53.1
B_P = 5.12 x 10^─8 T upward

Share :