Application of Ampere's Law : Magnetic Field Due to Circular Conductor

Q. A circular conductor of radius $r_o=1cm$ has an internal field,


$\overline{H}=\frac{{10}^4}{r_o}(\frac{1}{a^2}\sin a{r}_o-\frac{r_o}{a}\cos a{r}_o){\hat{a}}_{\varphi }\left(A/m\right)$

where, $a=\frac{\pi }{2r_0}$

1. $8\pi Ampere$
2. $\frac{8}{\pi }Ampere$
3. $4\pi Ampere$
4. $\frac{4}{\pi }Ampere$

Q. An infinitely long uniform solid wire of radius a carries a uniform dc current of density $\overrightarrow{j}.$

A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.

The magnetic field inside the hole is

1. uniform and depends only on d
2. uniform and depends only on b
3. uniform and depends on both b and d
4. non uniform

Q. The magnetic field intensity $\overrightarrow{H}$ at the center of the current of the carrying circular loops shown in the figure below is

1. $-\frac{I}{R}{\hat{a}}_z$
2. $-\frac{I}{2R}{\hat{a}}_z$
3. $\frac{I}{R}{\hat{a}}_z$
4. $\frac{I}{2R}{\hat{a}}_z$

Q. The current diensty in a medium is given by

$\overrightarrow{J}=\frac{400sin\theta }{2\pi (r^2+4)}{\hat{a}}_{r}A{m}^{-2}$

The total current and the average current density flowing through the portion of a spherical surface

r = 0.8 m, $\frac{\pi }{12}\le \theta \le \frac{\pi }{4},0\le 2\pi$ are given, respectively, by

1. 15.09 A, 12.86 $A{m}^{-2}$
2. 7.17 A, 6.88 $A{m}^{-2}$
3. 12.86 A, 9.23 $A{m}^{-2}$
4. 10.28 A, 7.56 $A{m}^{-2}$

Q. The magnetic field intensity $\overrightarrow{H}$ at the centre of current carrying loop (circular in space) and kept in x -y plane as shown will be

1. $\frac{I}{2R}\widehat{a_z}$
2. $-\frac{I}{2R}\widehat{a_z}$
3. $\frac{I}{R}\widehat{a_z}$
4. $-\frac{I}{R}\widehat{a_z}$