0.11 Ampere’s law (exercise)  (Page 2/4)

Currents and ampere’s loops

Solution :

According to Ampere’s law , the closed line integral is related to enclosed current as :

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

For loop 1, the integration direction is anticlockwise. The current out of the page is positive and current into the page is negative. There are one in and one out current here. The net current is zero. Hence,

$$\Rightarrow \oint \mathbf{B}.đ\mathbf{l}=0$$

For loop 2, the integration direction is clockwise. The current out of the page is negative and current into the page is positive. There are one in and two out current here. The net current is one out current i.e. “-I”. Hence,

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

For loop 3, the integration direction is clockwise. The current out of the page is negative and current into the page is positive. There are one in and one out current here. The net current is zero. Hence,

$$\oint \mathbf{B}.d\mathbf{l}=0$$

For loop 4, the integration direction is clockwise. The current out of the page is negative and current into the page is positive. There are three in and two out current here. The net current is one in current i.e. “I”. Hence,

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

Problem 4: The magnetic field in a region is given by relation :

$$\mathbf{B}=5\underset{}{\overset{‸}{\mathbf{l}}}$$

Closed line integral of magnetic field

where i is the unit vector in x direction. Determine closed line integral of magnetic field along the triangle ACD.

Solution :

For line segment AC, magnetic field and length element are in the same direction. Applying Ampere’s law :

$${\int }_{AC}\mathbf{B}.đ\mathbf{l}={\int }_{AC}5đx\cos 0°={\int }_{AC}5đx$$ $$\Rightarrow {\int }_{AC}\mathbf{B}.đ\mathbf{l}=5{\int }_{AC}đx=5X8=40Tm$$

We see that magnetic field is perpendicular to line segment CD. Therefore, magnetic line integral for this segment is equal to zero.

For the line segment DA, the length is $\sqrt{\left(8^2+6^2\right)}=10$ m.

Closed line integral of magnetic field

$${\int }_{DA}\mathbf{B}.đ\mathbf{l}={\int }_{DA}5đx\cos \angle AEF=5{\int }_{DA}đx\cos \left(\pi -\angle DEF\right)=-5{\int }_{DA}đx\cos \angle DEF$$ $$\Rightarrow {\int }_{DA}\mathbf{B}.đ\mathbf{l}=-5{\int }_{DA}đx\cos \angle DAC=-5{\int }_{DA}\frac{AC}{AD}đx=-5{\int }_{DA}\frac{8}{10}đx$$ $$\Rightarrow {\int }_{DA}\mathbf{B}.đ\mathbf{l}==-4{\int }_{DA}đx=-4XDA=-40Tm$$

Adding two values, the value of closed line integral is zero.

$$\oint \mathbf{B}.đ\mathbf{l}=0$$

Thus, we see that current though the region is zero even though there exists magnetic field in the region.

Problem 5: Straight wires are mounted tightly over a long hollow cylinder of radius "R" such that they are parallel to the axis of cylinder. The perpendicular cross-section of the arrangement is shown in the figure below. If there are N such wires each carrying a current I, then determine magnetic field inside and outside the cylinder.

Magnetic field due to current in tightly packed straight wires

Solution : We draw an Ampere loop of radius "r" for applying Ampere’s law at a point inside the cylinder. But there is no current inside. Hence,

Magnetic field due to current in tightly packed straight wires

$$\oint \mathbf{B}.đ\mathbf{l}=0$$

Here integration of dl along the loop is equal to perimeter of loop i.e. 2πr. Hence, B = 0. For determining magnetic field at an outside point, we draw an Ampere loop of radius "r". Here, the total current is NI. Hence,

$$\oint \mathbf{B}.đ\mathbf{l}={\mu }_{0}NI$$ $$\Rightarrow 2\pi rB={\mu }_{0}NI$$ $$\Rightarrow B=\frac{{\mu }_{0}NI}{2\pi r}$$