0.3 Magnetic field due to current in straight wire  (Page 2/4)

Magnetic field due to current in straight wire

Here ${\mathbf{B}}_1$ , ${\mathbf{B}}_2$ , ${\mathbf{B}}_3$ and ${\mathbf{B}}_4$ are the net magnetic fields at four different positions due to all current elements of the wire. If we draw a circular path around the straight wire such that its plane is perpendicular to the wire and its center lies on it, then each point on the perimeter is equidistant from the center. As such magnitudes of magnetic field on all points on the circle are equal. The direction of magnetic field as determined by right hand vector cross product rule is tangential to the circle.

Magnetic field due to current in straight wire

The observations as above are the basis of Right hand thumb rule for finding direction of magnetic field due to current in straight wire. If holding straight wire with right hand so that the extended thumb points in the direction of current, then curl of the fingers gives the direction of magnetic field around the straight wire.

Right hand thumb rule

Magnetic field due to current in finite straight wire

Since directions of magnetic fields due to all current elements are same, we can integrate the expression of magnitude as given by Biot-Savart law for the small current element (we have replaced dl by dy in accordance with notation in the figure) :

$$B=\int \mathrm{đB}=\frac{{\mu }_0}{4\pi }\int \frac{Iđy\sin \theta }{r^2}$$

Magnetic field due current in finite straight wire

In order to evaluate this integral in terms of angle φ, we determine đy, r and θ in terms of perpendicular distance “R” (which is a constant for a given point) and angle “φ”. Here,

$$y=R\tan \phi$$ $$dy=R{\sec }^2\phi đ\phi$$ $$r=R\sec \phi$$ $$\theta =\frac{\pi }{2}-\phi$$

Substituting in the integral, we have :

$$\Rightarrow B=\frac{{\mu }_0}{4\pi }\int \frac{IR{\sec }^2\phi đ\phi \sin \left(\frac{\pi }{2}-\phi \right)}{R^2{\sec }^2\phi }=\frac{{\mu }_0}{4\pi }\int \frac{I\cos \phi đ\phi }{R}$$

Taking out I and R out of the integral as they are constant :

$$\Rightarrow B=\frac{{\mu }_{0}I}{4\pi R}\int \cos \phi đ\phi$$

Integrating between angle “ ${\phi }_1$ ” and “ ${\phi }_2$ ”, we have :

$$\Rightarrow B=\frac{{\mu }_{0}I}{4\pi R}\underset{{\phi }_1}{\overset{{\phi }_2}{\int }}I\cos \phi đ\phi$$ $$\Rightarrow B=\frac{{\mu }_{0}I}{4\pi R}\left(\sin {\phi }_2-\sin {\phi }_1\right)$$

We follow the convention whereby angle is measured from perpendicular line. The angle below perpendicular line is treated negative and angle above perpendicular line is positive. In case, we want to do away with the sign of angle, we put ${\phi }_1=-{\phi }_1$ in above equation :

$$\Rightarrow B=\frac{{\mu }_{0}I}{4\pi R}\left(\sin {\phi }_1+\sin {\phi }_2\right)$$

Note that angles being used with this expression are positive numbers only. Also note that the magnitude of magnetic field depends on where the point of observation P lies with respect to straight wire, which is reflected in the value of angle φ.

We can also express magnetic field due to current in a straight wire at a perpendicular distance “R” in terms of angles between straight wire and line joining point of observation and end points.

Magnetic field due to current in wire

$$B=\frac{{\mu }_{0}I}{4\pi R}X[\sin \left(\frac{\pi }{2}-{\theta }_1\right)+\sin \left(\frac{\pi }{2}-{\theta }_2\right)]$$ $$\Rightarrow B=\frac{{\mu }_{0}I}{4\pi R}X[\cos {\theta }_1+\cos {\theta }_2]$$