Question 12: Magnetism and Matter

Question

An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

Accepted Answer

The segments AB and DE together form an infinitely long straight wire.
For an infinite wire the magnitude of the magnetic field at distance r is given as
$$ \mathrm{B}_{	ext {straight }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} $$
Using the Right-Hand Thumb Rule, if the thumb points in the direction of current I (along +y ), the fingers curl out of the page at point O , i.e., in the -x direction.
$$ \overrightarrow{\mathrm{B}}_{	ext {straight }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \hat{\imath} $$
The circular part is a nearly complete loop of radius r .
For a circular loop at its centre the magnitude of the magnetic field is given as,
$$ \mathrm{B}_{	ext {loop }}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}} $$
Using the Right-Hand Rule for loops, curling the fingers in the direction of the current (clockwise in the yz -plane) the thumb points into the page, which is the -x direction.
$$ \overrightarrow{\mathrm{B}}_{\mathrm{loop}}=-\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}} \hat{\mathrm{i}} $$
The total field is the vector sum of both parts :
$$ \overrightarrow{\mathrm{B}}_{	ext {total }}=\overrightarrow{\mathrm{B}}_{	ext {straight }}+\overrightarrow{\mathrm{B}}_{\mathrm{loop}} $$
$$\Rightarrow $$ $\overrightarrow{\mathrm{B}}_{\mathrm{total}}=\left(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}ight) \hat{\mathrm{i}}+\left(-\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}ight) \hat{\mathrm{i}}$
$$\Rightarrow $$ $\overrightarrow{\mathrm{B}}_{	ext {total }}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left[\frac{1}{\pi}-1ight] \hat{\mathrm{i}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}[1-\pi] \hat{\mathrm{i}}=-\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}[\pi-1] \hat{\mathrm{i}}$
So, the magnetic field at the centre O of the circular loop is $=-\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}[\pi-1] \hat{1}$.
Hence, the correct option is (B).

Answer

$-\frac{\mu_0}{2\pi} \frac{I}{r} (\pi + 1) \hat{i}$

Answer

$-\frac{\mu_0}{2\pi} \frac{I}{r} (\pi - 1) \hat{i}$

Answer

$\frac{\mu_0}{2\pi} \frac{I}{r} (\pi + 1) \hat{i}$

Answer

$\frac{\mu_0}{2\pi} \frac{I}{r} (\pi - 1) \hat{i}$

Magnetism and Matter

An infinitely long straight wire carrying current I is bent in a planar shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :

Choose the correct answer: