0.11 Ampere’s law (exercise)  (Page 3/4)

Problem 6: A cylindrical conductor of radius R carries current I distributed uniformly across the cross-section. Draw the curve showing variation of magnetic field as we move away from the axis of conductor in perpendicular direction.

Solution : Let the perpendicular direction to the axis be x-axis. The magnetic field at a point inside the conductor is given by :

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

Clearly, magnetic field increases linearly as move away from the axis towards the edge of conductor and attains the maximum at the surface, when x=R and magnetic field is given as:

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

The magnetic field at a point outside the conductor is :

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

The magnetic field is inversely proportional to the linear distance “x”. The required plot of magnetic field .vs. x is as shown in the figure below :

Variation of magnetic field

Problem 7: A long annular cylindrical conductor of radii “a” and “b” carries current I. The perpendicular cross section of annular cylinder is shown in the figure below. If the current distribution in the annular region is uniform, determine magnetic field at a point in the annular region at a radial distance “r” from the axis.

Magnetic field due to current in annular cylindrical conductor

Solution : According to Ampere’s law ,

$$\oint \mathbf{B}.đ\mathbf{l}={\mu }_{0}I$$

In order to evaluate this equation, we need to know the current in the annular region from r=a to r=r. For this we need the value of current density. Here, total current is given. Dividing this by the total area of the region gives us the current density,

$$J=\frac{I}{\pi \left(b^2-a^2\right)}$$

The net current through the Ampere loop of radius “r” falling in the annular region is given by multiplying current density with the annular area between r=a and r=r. Applying Ampere’s law for a loop of radius r,

Magnetic field due to current in annular cylindrical conductor

$$\oint \mathbf{B}.đ\mathbf{l}={\mu }_0\pi \left(r^2-a^2\right)J=\frac{{\mu }_0\pi \left(r^2-a^2\right)I}{\pi \left(b^2-a^2\right)}$$ $$\Rightarrow 2\pi rB=\frac{{\mu }_{0}I\left(r^2-a^2\right)}{\left(b^2-a^2\right)}$$ $$\Rightarrow B=\frac{{\mu }_{0}I\left(r^2-a^2\right)}{2\pi r\left(b^2-a^2\right)};a<r<b$$

Problem 8: A long annular cylindrical conductor of radii “a” and “b” carries a current. If the current distribution in the annular region is given as J = kr, where k is a constant, then determine magnetic field at a point in the annular region at a radial distance “r” from the axis.

Solution : This question is similar to earlier question with one difference that areal current density is not uniform. We see here that the current distribution in the annular region is given as J=kr. Clearly, current density increases as we move from inner edge to the outer edge of the annular cylinder. The current in the small strip đr is :

$$đI=2\pi rđrJ=2\pi rđrkr=2\pi k{r}^2đr$$

Magnetic field due to current in annular cylindrical conductor

Applying Ampere’s law for a loop of radius r and considering that current is distributed from r=a to r=r,

$$\oint \mathbf{B}.d\mathbf{l}={\mu }_0\oint dI={\mu }_0\underset{a}{\overset{r}{\int }}2\pi k{r}^{2}dr=2\pi {\mu }_{0}k\underset{a}{\overset{r}{\int }}{r}^{2}dr$$ $$\Rightarrow 2\pi rB=2\pi {\mu }_{0}k[\frac{r^3}{3}\underset{a}{\overset{r}{]}}=\frac{2\pi {\mu }_{0}k\left(r^3-a^3\right)}{3}$$ $$\Rightarrow B=\frac{{\mu }_{0}k\left(r^3-a^3\right)}{3r};a<r<b$$

Problem 9: A long solenoid having 1000 turns per meter carries a current of 1 A. A long straight conductor of radius 0.5 cm and carrying a current of 10π A is placed coaxially along the axis of solenoid. Compare magnetic fields due to two currents at that point. Also determine magnetic field at a point on the surface of straight conductor.