Chapter 29: Q13P (page 857)

In Fig. 29-44 point $P_{1}$ is at distance $R = 13.1 cm$ on the perpendicular bisector of a straight wire of length $L = 18.0 cm$ . carrying current. (Note that the wire is notlong.) What is the magnitude of the magnetic field at $P_{1}$ due to i?

Short Answer

Expert verified

The magnitude of the magnetic field at $P_{1}$ due to i is $5.03 \times 10^{- 8} T$

Step by step solution

Given

Current $i = 58.2 mA$
Distance of point P on the perpendicular bisector of wire $R = 13.1 cm = 0.131 m$
Length of wire $L = 18.0 cm = 0.18 m$

Determine the formulas:

Use the formula of Biot-Savarts law to find the magnitude of the field at $P_{1}$ due to i

Formula:

$d B = \frac{μ_{0} i d l}{4 π} \frac{s i n θ}{r^{2}}$

Calculate the magnitude of the magnetic field at P1 due to i

According to Bio-Savarts law

$d B = \frac{μ_{0} i d l}{4 π} \frac{\sin θ}{r^{2}}$

If r makes an angle $θ$ withl then

$r = \sqrt{l^{2} + R^{2}}$

And

$\sin θ = \frac{R}{R} = \frac{R}{\sqrt{l^{2} + R^{2}}} l = \frac{L}{2} = \frac{0.18 m}{2} = 0.09 m$

Integrating an equation of Bio-Savarts law

$B = \int_{0}^{0.09} d B = \int_{0}^{0.09} \frac{μ_{0} i d l}{4 π} \frac{\sin θ}{r^{2}} B = \frac{μ_{0} i}{4 π} \int_{0}^{0.09} \frac{\sin θ d l}{r^{2}} B = \frac{μ_{0} i}{4 π} \int_{0}^{0.09} \frac{R}{\sqrt{l^{2} + R^{2}}} \frac{d l}{{(\sqrt{l^{2} + R^{2}})}^{2}} B = \frac{μ_{0} i R}{4 π} \int_{0}^{0.09} \frac{d l}{{(l^{2} + R^{2})}^{3 / 2}}$

Solve further as:

$B = \frac{μ_{0} i R}{4 π} \frac{1}{R^{2}} {|\frac{l}{{(l^{2} + R^{2})}^{1 / 2}}|}_{0}^{0.09} B = \frac{μ_{0} i}{2 π R} [\frac{l}{{(l^{2} + R^{2})}^{1 / 2}}] B = \frac{(4 π \times 10^{- 7} \frac{Tm}{A}) (0.0582 A)}{2 π (0.131 m)} [\frac{0.09 m}{{[{(0.09 m)}^{2} + {(0.131 m)}^{2}]}^{1 / 2}}] B = 5.03 \times 10^{- 8} T$