153192
Find magnetic field at centre $P$ if length of side of square loop is $20 \mathrm{~cm}$.
1 $12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
2 $12 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $6 \sqrt{2} \times 10^{-6} \mathrm{~T}$
Explanation:
A Given that, $\mathrm{I}=3 \mathrm{~A}, \mathrm{a}=20 \mathrm{~cm}=0.2 \mathrm{~m}, \mathrm{EP}=\mathrm{r}=\frac{\mathrm{a}}{2}$ Each side of square will act as a conductor of length a. Since, we know that magnetic field due to current carrying conductor at a distance $r$ - $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{r}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ So, magnetic field due to all sides of square at its center- $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}}\left[\sin \theta_{1}+\sin \theta_{2}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{(\mathrm{a} / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}=2 \sqrt{2} \frac{\mu_{0} \mathrm{I}}{\pi \mathrm{a}}$ $\mathrm{B}=2 \sqrt{2} \times \frac{4 \pi \times 10^{-7} \times 3}{\pi \times 0.2}$ $\mathrm{~B}=12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
AIIMS-26.05.2019(M) Shift-1
Moving Charges & Magnetism
153193
In figure two parallel infinitely long current carrying wires are shown. If resultant magnetic field at point $\mathrm{A}$ is zero. Then determine current $\mathrm{I}$.
1 $50 \mathrm{~A}$
2 $15 \mathrm{~A}$
3 $30 \mathrm{~A}$
4 $25 \mathrm{~A}$
Explanation:
C Given that, $I_{1}=10 A, I_{2}=I, r_{2}=27 \mathrm{~cm}, r_{1}=9 \mathrm{~cm}$, Net magnetic field at point $A$ - $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}\left(\because \mathrm{I}_{1}\right.$ and $\mathrm{I}_{2}$ are opposite in direction $)$ According to question- $\mathrm{B}_{\text {net }}=0$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ $\frac{\mu_{0} I_{1}}{2 \pi r_{1}}=\frac{\mu_{0} I_{2}}{2 \pi r_{2}}$ $I_{2}=I_{1} \times \frac{r_{2}}{r_{1}}, \quad I=10 \times \frac{27}{9}=30 A$
AIIMS-25.05.2019
Moving Charges & Magnetism
153194
A semi circular arc of radius $r$ and a straight wire along the diameter, both are carrying same current $i$. Find out magnetic force per unit length at point $P$ at centre.
1 $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$
2 $\left(\frac{\mu_{0} i^{2}}{2 r}\right)$
3 $\left(\frac{\mu_{0} i^{2}}{r}\right)$
4 $\left(\frac{2 \mu_{0} i^{2}}{r}\right)$
Explanation:
A The magnetic force per unit length $\mathrm{F}=\mathrm{Bi} l$ $\frac{\mathrm{F}}{l}=\mathrm{Bi}$ $=\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}} \times \mathrm{i}$ $\left(\frac{\mathrm{F}}{l}\right)=\frac{\mu_{0} \mathrm{i}^{2}}{4 \mathrm{r}}$ The magnetic force per unit length is $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$.
AIIMS-25.05.2019(M) Shift-1
Moving Charges & Magnetism
153195
In the given figure, find out magnetic field at point \(B\) (Given: \(I=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}\) )
A Given $\mathrm{I}=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Net magnetic field at point $\mathrm{B}$, $\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\text {ring }}+\overrightarrow{\mathrm{B}}_{\text {wire }}$ Since, we know that magnetic field due to wire and ring at a point $\mathrm{B}$ is given as- $\overrightarrow{\mathrm{B}}_{\text {ring }}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\overrightarrow{\mathrm{B}}_{\text {wire }}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\overrightarrow{\mathrm{B}}=\left(\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}+\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\right)$ $\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\frac{4 \pi \times 10^{-7} \times 2.5}{2 \times 5 \times 10^{-2}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\pi \times\left[1+\frac{1}{\pi}\right] \times 10^{-5} \mathrm{~T}$
153192
Find magnetic field at centre $P$ if length of side of square loop is $20 \mathrm{~cm}$.
1 $12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
2 $12 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $6 \sqrt{2} \times 10^{-6} \mathrm{~T}$
Explanation:
A Given that, $\mathrm{I}=3 \mathrm{~A}, \mathrm{a}=20 \mathrm{~cm}=0.2 \mathrm{~m}, \mathrm{EP}=\mathrm{r}=\frac{\mathrm{a}}{2}$ Each side of square will act as a conductor of length a. Since, we know that magnetic field due to current carrying conductor at a distance $r$ - $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{r}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ So, magnetic field due to all sides of square at its center- $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}}\left[\sin \theta_{1}+\sin \theta_{2}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{(\mathrm{a} / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}=2 \sqrt{2} \frac{\mu_{0} \mathrm{I}}{\pi \mathrm{a}}$ $\mathrm{B}=2 \sqrt{2} \times \frac{4 \pi \times 10^{-7} \times 3}{\pi \times 0.2}$ $\mathrm{~B}=12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
AIIMS-26.05.2019(M) Shift-1
Moving Charges & Magnetism
153193
In figure two parallel infinitely long current carrying wires are shown. If resultant magnetic field at point $\mathrm{A}$ is zero. Then determine current $\mathrm{I}$.
1 $50 \mathrm{~A}$
2 $15 \mathrm{~A}$
3 $30 \mathrm{~A}$
4 $25 \mathrm{~A}$
Explanation:
C Given that, $I_{1}=10 A, I_{2}=I, r_{2}=27 \mathrm{~cm}, r_{1}=9 \mathrm{~cm}$, Net magnetic field at point $A$ - $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}\left(\because \mathrm{I}_{1}\right.$ and $\mathrm{I}_{2}$ are opposite in direction $)$ According to question- $\mathrm{B}_{\text {net }}=0$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ $\frac{\mu_{0} I_{1}}{2 \pi r_{1}}=\frac{\mu_{0} I_{2}}{2 \pi r_{2}}$ $I_{2}=I_{1} \times \frac{r_{2}}{r_{1}}, \quad I=10 \times \frac{27}{9}=30 A$
AIIMS-25.05.2019
Moving Charges & Magnetism
153194
A semi circular arc of radius $r$ and a straight wire along the diameter, both are carrying same current $i$. Find out magnetic force per unit length at point $P$ at centre.
1 $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$
2 $\left(\frac{\mu_{0} i^{2}}{2 r}\right)$
3 $\left(\frac{\mu_{0} i^{2}}{r}\right)$
4 $\left(\frac{2 \mu_{0} i^{2}}{r}\right)$
Explanation:
A The magnetic force per unit length $\mathrm{F}=\mathrm{Bi} l$ $\frac{\mathrm{F}}{l}=\mathrm{Bi}$ $=\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}} \times \mathrm{i}$ $\left(\frac{\mathrm{F}}{l}\right)=\frac{\mu_{0} \mathrm{i}^{2}}{4 \mathrm{r}}$ The magnetic force per unit length is $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$.
AIIMS-25.05.2019(M) Shift-1
Moving Charges & Magnetism
153195
In the given figure, find out magnetic field at point \(B\) (Given: \(I=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}\) )
A Given $\mathrm{I}=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Net magnetic field at point $\mathrm{B}$, $\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\text {ring }}+\overrightarrow{\mathrm{B}}_{\text {wire }}$ Since, we know that magnetic field due to wire and ring at a point $\mathrm{B}$ is given as- $\overrightarrow{\mathrm{B}}_{\text {ring }}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\overrightarrow{\mathrm{B}}_{\text {wire }}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\overrightarrow{\mathrm{B}}=\left(\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}+\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\right)$ $\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\frac{4 \pi \times 10^{-7} \times 2.5}{2 \times 5 \times 10^{-2}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\pi \times\left[1+\frac{1}{\pi}\right] \times 10^{-5} \mathrm{~T}$
153192
Find magnetic field at centre $P$ if length of side of square loop is $20 \mathrm{~cm}$.
1 $12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
2 $12 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $6 \sqrt{2} \times 10^{-6} \mathrm{~T}$
Explanation:
A Given that, $\mathrm{I}=3 \mathrm{~A}, \mathrm{a}=20 \mathrm{~cm}=0.2 \mathrm{~m}, \mathrm{EP}=\mathrm{r}=\frac{\mathrm{a}}{2}$ Each side of square will act as a conductor of length a. Since, we know that magnetic field due to current carrying conductor at a distance $r$ - $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{r}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ So, magnetic field due to all sides of square at its center- $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}}\left[\sin \theta_{1}+\sin \theta_{2}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{(\mathrm{a} / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}=2 \sqrt{2} \frac{\mu_{0} \mathrm{I}}{\pi \mathrm{a}}$ $\mathrm{B}=2 \sqrt{2} \times \frac{4 \pi \times 10^{-7} \times 3}{\pi \times 0.2}$ $\mathrm{~B}=12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
AIIMS-26.05.2019(M) Shift-1
Moving Charges & Magnetism
153193
In figure two parallel infinitely long current carrying wires are shown. If resultant magnetic field at point $\mathrm{A}$ is zero. Then determine current $\mathrm{I}$.
1 $50 \mathrm{~A}$
2 $15 \mathrm{~A}$
3 $30 \mathrm{~A}$
4 $25 \mathrm{~A}$
Explanation:
C Given that, $I_{1}=10 A, I_{2}=I, r_{2}=27 \mathrm{~cm}, r_{1}=9 \mathrm{~cm}$, Net magnetic field at point $A$ - $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}\left(\because \mathrm{I}_{1}\right.$ and $\mathrm{I}_{2}$ are opposite in direction $)$ According to question- $\mathrm{B}_{\text {net }}=0$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ $\frac{\mu_{0} I_{1}}{2 \pi r_{1}}=\frac{\mu_{0} I_{2}}{2 \pi r_{2}}$ $I_{2}=I_{1} \times \frac{r_{2}}{r_{1}}, \quad I=10 \times \frac{27}{9}=30 A$
AIIMS-25.05.2019
Moving Charges & Magnetism
153194
A semi circular arc of radius $r$ and a straight wire along the diameter, both are carrying same current $i$. Find out magnetic force per unit length at point $P$ at centre.
1 $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$
2 $\left(\frac{\mu_{0} i^{2}}{2 r}\right)$
3 $\left(\frac{\mu_{0} i^{2}}{r}\right)$
4 $\left(\frac{2 \mu_{0} i^{2}}{r}\right)$
Explanation:
A The magnetic force per unit length $\mathrm{F}=\mathrm{Bi} l$ $\frac{\mathrm{F}}{l}=\mathrm{Bi}$ $=\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}} \times \mathrm{i}$ $\left(\frac{\mathrm{F}}{l}\right)=\frac{\mu_{0} \mathrm{i}^{2}}{4 \mathrm{r}}$ The magnetic force per unit length is $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$.
AIIMS-25.05.2019(M) Shift-1
Moving Charges & Magnetism
153195
In the given figure, find out magnetic field at point \(B\) (Given: \(I=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}\) )
A Given $\mathrm{I}=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Net magnetic field at point $\mathrm{B}$, $\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\text {ring }}+\overrightarrow{\mathrm{B}}_{\text {wire }}$ Since, we know that magnetic field due to wire and ring at a point $\mathrm{B}$ is given as- $\overrightarrow{\mathrm{B}}_{\text {ring }}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\overrightarrow{\mathrm{B}}_{\text {wire }}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\overrightarrow{\mathrm{B}}=\left(\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}+\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\right)$ $\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\frac{4 \pi \times 10^{-7} \times 2.5}{2 \times 5 \times 10^{-2}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\pi \times\left[1+\frac{1}{\pi}\right] \times 10^{-5} \mathrm{~T}$
153192
Find magnetic field at centre $P$ if length of side of square loop is $20 \mathrm{~cm}$.
1 $12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
2 $12 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $6 \sqrt{2} \times 10^{-6} \mathrm{~T}$
Explanation:
A Given that, $\mathrm{I}=3 \mathrm{~A}, \mathrm{a}=20 \mathrm{~cm}=0.2 \mathrm{~m}, \mathrm{EP}=\mathrm{r}=\frac{\mathrm{a}}{2}$ Each side of square will act as a conductor of length a. Since, we know that magnetic field due to current carrying conductor at a distance $r$ - $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{r}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ So, magnetic field due to all sides of square at its center- $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}}\left[\sin \theta_{1}+\sin \theta_{2}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{(\mathrm{a} / 2)}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}=4 \times \frac{\mu_{0}}{4 \pi} \times \frac{2 \mathrm{I}}{\mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}=2 \sqrt{2} \frac{\mu_{0} \mathrm{I}}{\pi \mathrm{a}}$ $\mathrm{B}=2 \sqrt{2} \times \frac{4 \pi \times 10^{-7} \times 3}{\pi \times 0.2}$ $\mathrm{~B}=12 \sqrt{2} \times 10^{-6} \mathrm{~T}$
AIIMS-26.05.2019(M) Shift-1
Moving Charges & Magnetism
153193
In figure two parallel infinitely long current carrying wires are shown. If resultant magnetic field at point $\mathrm{A}$ is zero. Then determine current $\mathrm{I}$.
1 $50 \mathrm{~A}$
2 $15 \mathrm{~A}$
3 $30 \mathrm{~A}$
4 $25 \mathrm{~A}$
Explanation:
C Given that, $I_{1}=10 A, I_{2}=I, r_{2}=27 \mathrm{~cm}, r_{1}=9 \mathrm{~cm}$, Net magnetic field at point $A$ - $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}\left(\because \mathrm{I}_{1}\right.$ and $\mathrm{I}_{2}$ are opposite in direction $)$ According to question- $\mathrm{B}_{\text {net }}=0$ $\therefore \quad \mathrm{B}_{1}=\mathrm{B}_{2}$ $\frac{\mu_{0} I_{1}}{2 \pi r_{1}}=\frac{\mu_{0} I_{2}}{2 \pi r_{2}}$ $I_{2}=I_{1} \times \frac{r_{2}}{r_{1}}, \quad I=10 \times \frac{27}{9}=30 A$
AIIMS-25.05.2019
Moving Charges & Magnetism
153194
A semi circular arc of radius $r$ and a straight wire along the diameter, both are carrying same current $i$. Find out magnetic force per unit length at point $P$ at centre.
1 $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$
2 $\left(\frac{\mu_{0} i^{2}}{2 r}\right)$
3 $\left(\frac{\mu_{0} i^{2}}{r}\right)$
4 $\left(\frac{2 \mu_{0} i^{2}}{r}\right)$
Explanation:
A The magnetic force per unit length $\mathrm{F}=\mathrm{Bi} l$ $\frac{\mathrm{F}}{l}=\mathrm{Bi}$ $=\frac{\mu_{0} \mathrm{i}}{4 \mathrm{r}} \times \mathrm{i}$ $\left(\frac{\mathrm{F}}{l}\right)=\frac{\mu_{0} \mathrm{i}^{2}}{4 \mathrm{r}}$ The magnetic force per unit length is $\left(\frac{\mu_{0} i^{2}}{4 r}\right)$.
AIIMS-25.05.2019(M) Shift-1
Moving Charges & Magnetism
153195
In the given figure, find out magnetic field at point \(B\) (Given: \(I=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}\) )
A Given $\mathrm{I}=2.5 \mathrm{~A}, \mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Net magnetic field at point $\mathrm{B}$, $\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\text {ring }}+\overrightarrow{\mathrm{B}}_{\text {wire }}$ Since, we know that magnetic field due to wire and ring at a point $\mathrm{B}$ is given as- $\overrightarrow{\mathrm{B}}_{\text {ring }}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ $\overrightarrow{\mathrm{B}}_{\text {wire }}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\overrightarrow{\mathrm{B}}=\left(\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}+\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\right)$ $\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\frac{4 \pi \times 10^{-7} \times 2.5}{2 \times 5 \times 10^{-2}}\left[1+\frac{1}{\pi}\right]$ $\overrightarrow{\mathrm{B}}=\pi \times\left[1+\frac{1}{\pi}\right] \times 10^{-5} \mathrm{~T}$
