0.11 Ampere’s law (exercise)

Worked out exercises

Problem 1: Two wires each carrying current I are perpendicular to xy plane. The current in one of them is into the plane denoted by a cross sign and the current in the other wire is out of the plane denoted by a filled circle. If the linear distance between the positions of two wires is “2a”, then find the net magnetic field at a distance”b” on the perpendicular bisector of the line joining the positions of two wires.

Magnetic field at perpendicular bisector

Solution : The magnitudes of magnetic fields due to wires at A and B are equal. Applying Ampere's law, the magnetic field due to each wire is :

$$B=\frac{{\mu }_{0}I}{2\pi r}$$

The magnetic fields are directed tangential to the circle drawn containing point “P” with centers “A” and “B” as shown in the figure. Each magnetic field makes an angle say “θ” with the bisector. The components in y-direction cancel out, whereas x-components add up. Clearly, the net magnetic field is directed in negative x – direction. The magnitude of net magnetic field is :

Magnetic field at perpendicular bisector

$$\Rightarrow B=2X\frac{{\mu }_{0}I\cos \theta }{2\pi r}=\frac{{\mu }_{0}I\cos \theta }{\pi r}$$

Now,

$$\cos \theta =\frac{a}{r}=\frac{a}{\sqrt{\left(a^2+b^2\right)}}$$

and

$$r=\sqrt{\left(a^2+b^2\right)}$$

Putting these expressions in the equation for the magnetic field at “P”, we have :

$$\Rightarrow B=\frac{{\mu }_{0}I\cos \theta }{\pi r}=\frac{{\mu }_{0}Ia}{\pi \sqrt{\left(a^2+b^2\right)}\sqrt{\left(a^2+b^2\right)}}$$ $$\Rightarrow B=\frac{{\mu }_{0}Ia}{\pi \left(a^2+b^2\right)}$$

Problem 2: Five straight wires, carrying current I, are perpendicular to the plane of drawing. Four of them are situated at the corners and fifth wire is situated at the center of a square of side "a". Two of the wires at the corners are flowing into the plane whereas the remaining three are flowing out of the plane. Find the net magnetic field at the center of square.

Five straight wires, carrying current i

Solution : According to Ampere’s law , the magnetic field due to a straight wire carrying current "I" at a perpendicular distance "r" is given as :

$$B=\frac{{\mu }_{0}I}{2\pi R}$$

The wires at the corners carry equal currents and the center "O" is equidistant from these wires. Thus, magnetic fields due to these four wires have equal magnitude. In order to find the directions of magnetic fields, we draw circles containing point of observation "O". The direction of magnetic field is tangential to the circle. Applying Right hand thumb rule for straight wire, we determine the orientation of magnetic field as shown in the figure. Clearly, the net magnetic field due to these four wires at the center is zero.

Directions of magnetic fields

Now, magnetic field at a point on the wire itself is zero. Thus, magnetic fields due to all the five wires at the center "O" is zero.

It is interesting to note that if straight wires with currents are arranged differently, for example, two currents out of the plane at A and C respectively and the other two currents into the plane at D and E respectively are arranged, then magnetic fields do not cancel and there is net non-zero magnetic field at "O" due to currents in four wires.

Problem 3: There are five long wires perpendicular to the plane of drawing, each carrying current I as shown by filled circles (out of plane) and crosses (into the plane) in the figure below. Determine closed line integrals $\oint \mathbf{B}.d\mathbf{l}$ for each of the four contours in the direction of integration shown.