153394
A current of $1 \mathrm{~A}$ is flowing along the sides of an equilateral triangle of side $4.5 \times 10^{-2} \mathrm{~m}$. The magnetic field at the centroid of the triangle is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)$
1 $4 \times 10^{-5} \mathrm{~T}$
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-4} \mathrm{~T}$
4 $2 \times 10^{-4} \mathrm{~T}$
Explanation:
A Given, $l=4.5 \times 10^{-2} \mathrm{~m}, \mathrm{I}=1 \mathrm{~A}$ Total magnetic field at the centroid due to all sides of a triangle is given by: $\mathrm{B}=\frac{3 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ For an equilateral triangle \(\theta_1=\theta_2=60^{\circ}\) From $\triangle \mathrm{BOD}$, $\tan 60^{\circ}=\mathrm{BD} / \mathrm{OD}$ $\mathrm{OD}=\mathrm{a}=\mathrm{BD} / \tan 60^{\circ}=\frac{l / 2}{\sqrt{3}} \quad(\because \mathrm{BD}=l / 2)$ $\mathrm{a}=\frac{l}{2 \sqrt{3}}=\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}$ $\therefore \quad \mathrm{B}=\frac{3 \times 4 \pi \times 10^{-7} \times 1 \times 2 \sqrt{3}}{4 \pi \times 4.5 \times 10^{-2}}\left(\frac{2 \sqrt{3}}{2}\right)$ $\mathrm{B}=4 \times 10^{-5} \mathrm{~T}$
TS EAMCET (Engg.)-2016
Moving Charges & Magnetism
153395
A square loop is made by a uniform conductor wire as shown in figure The net magnetic field at the centre of the loop if side length of the square is a
1 $\frac{\mu_{0} \mathrm{i}}{2 \mathrm{a}}$
2 Zero
3 $\frac{\mu_{0} i^{2}}{a^{2}}$
4 None of these
Explanation:
B From the figure, we can see that both part PS-R and P-Q-R have the same length so, current in both branches will be same and using the Biot-Savart law the field at the center is zero.
JIPMER-2015
Moving Charges & Magnetism
153397
A long straight wire of radius $R$ carries a steady current $I$. The current is uniformly distributed across its cross-section. The ratio of magnetic field at $R / 2$ and $2 R$ is
1 $\frac{1}{2}$
2 2
3 $\frac{1}{4}$
4 1
Explanation:
D Consider two amperion loops of radius R/2 Ans: d : Consider two amperion loops of radius $R / 2$ and 2R. Applying ampere circular law for their loops we get- For a point inside the long straight wire $\mathrm{B}_{\text {inside }}=\frac{\mu_{0} \mathrm{ir}}{2 \pi \mathrm{a}^{2}}$ $\text { at } \mathrm{r}=\frac{\mathrm{a}}{2}$ $\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ For a point out side the solid cylinder $\mathrm{B}_{\text {out }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\text { at } \mathrm{r}=2 \mathrm{a}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}} \times \frac{4 \pi \mathrm{a}}{\mu_{0} \mathrm{i}}=1$
AMU-2016
Moving Charges & Magnetism
153398
Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What potential difference ratio should be applied across them so that the magnetic field at their centers is the same?
1 2
2 3
3 4
4 6
Explanation:
C Magnetic field due to $1^{\text {st }}$ coil and second coil same at center. Then, magnetic field $\mathrm{B}$ at center is, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}_{1}}{2(2 \mathrm{r})}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \mathrm{r}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2$ As we know that, resistance of coil is related as $\mathrm{R}=\rho \frac{1}{\mathrm{~A}} \quad \text { where, } \rho =\text { resistivity }$ $1 =\text { Length }$ $\mathrm{A} =\text { Area of cross }$ $\rho$ and $A$ is same for both coil but $I_{1}=2 \pi(2 r)$ $\text { And } \quad 1_{2}=2 \pi(\mathrm{r})$ If $V_{1}$, and $V_{2}$ applied voltage first and second coil then, $I_{1}=\frac{V_{1}}{R_{1}} \text { and } I_{2}=\frac{V_{2}}{R_{2}}$ $I_{1}=\frac{V_{1}}{\rho \frac{l_{1}}{A}}$ $I_{2}=\frac{V_{2}}{\rho \frac{l_{2}}{A}}$ From equation (i), (ii) and (iii)- $\frac{\mathrm{V}_{1}}{l_{1}} \times \frac{l_{2}}{\mathrm{~V}_{2}}=2, \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{l_{2}}{l_{1}}=2, \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{1}{2}=2$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$
153394
A current of $1 \mathrm{~A}$ is flowing along the sides of an equilateral triangle of side $4.5 \times 10^{-2} \mathrm{~m}$. The magnetic field at the centroid of the triangle is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)$
1 $4 \times 10^{-5} \mathrm{~T}$
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-4} \mathrm{~T}$
4 $2 \times 10^{-4} \mathrm{~T}$
Explanation:
A Given, $l=4.5 \times 10^{-2} \mathrm{~m}, \mathrm{I}=1 \mathrm{~A}$ Total magnetic field at the centroid due to all sides of a triangle is given by: $\mathrm{B}=\frac{3 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ For an equilateral triangle \(\theta_1=\theta_2=60^{\circ}\) From $\triangle \mathrm{BOD}$, $\tan 60^{\circ}=\mathrm{BD} / \mathrm{OD}$ $\mathrm{OD}=\mathrm{a}=\mathrm{BD} / \tan 60^{\circ}=\frac{l / 2}{\sqrt{3}} \quad(\because \mathrm{BD}=l / 2)$ $\mathrm{a}=\frac{l}{2 \sqrt{3}}=\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}$ $\therefore \quad \mathrm{B}=\frac{3 \times 4 \pi \times 10^{-7} \times 1 \times 2 \sqrt{3}}{4 \pi \times 4.5 \times 10^{-2}}\left(\frac{2 \sqrt{3}}{2}\right)$ $\mathrm{B}=4 \times 10^{-5} \mathrm{~T}$
TS EAMCET (Engg.)-2016
Moving Charges & Magnetism
153395
A square loop is made by a uniform conductor wire as shown in figure The net magnetic field at the centre of the loop if side length of the square is a
1 $\frac{\mu_{0} \mathrm{i}}{2 \mathrm{a}}$
2 Zero
3 $\frac{\mu_{0} i^{2}}{a^{2}}$
4 None of these
Explanation:
B From the figure, we can see that both part PS-R and P-Q-R have the same length so, current in both branches will be same and using the Biot-Savart law the field at the center is zero.
JIPMER-2015
Moving Charges & Magnetism
153397
A long straight wire of radius $R$ carries a steady current $I$. The current is uniformly distributed across its cross-section. The ratio of magnetic field at $R / 2$ and $2 R$ is
1 $\frac{1}{2}$
2 2
3 $\frac{1}{4}$
4 1
Explanation:
D Consider two amperion loops of radius R/2 Ans: d : Consider two amperion loops of radius $R / 2$ and 2R. Applying ampere circular law for their loops we get- For a point inside the long straight wire $\mathrm{B}_{\text {inside }}=\frac{\mu_{0} \mathrm{ir}}{2 \pi \mathrm{a}^{2}}$ $\text { at } \mathrm{r}=\frac{\mathrm{a}}{2}$ $\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ For a point out side the solid cylinder $\mathrm{B}_{\text {out }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\text { at } \mathrm{r}=2 \mathrm{a}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}} \times \frac{4 \pi \mathrm{a}}{\mu_{0} \mathrm{i}}=1$
AMU-2016
Moving Charges & Magnetism
153398
Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What potential difference ratio should be applied across them so that the magnetic field at their centers is the same?
1 2
2 3
3 4
4 6
Explanation:
C Magnetic field due to $1^{\text {st }}$ coil and second coil same at center. Then, magnetic field $\mathrm{B}$ at center is, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}_{1}}{2(2 \mathrm{r})}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \mathrm{r}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2$ As we know that, resistance of coil is related as $\mathrm{R}=\rho \frac{1}{\mathrm{~A}} \quad \text { where, } \rho =\text { resistivity }$ $1 =\text { Length }$ $\mathrm{A} =\text { Area of cross }$ $\rho$ and $A$ is same for both coil but $I_{1}=2 \pi(2 r)$ $\text { And } \quad 1_{2}=2 \pi(\mathrm{r})$ If $V_{1}$, and $V_{2}$ applied voltage first and second coil then, $I_{1}=\frac{V_{1}}{R_{1}} \text { and } I_{2}=\frac{V_{2}}{R_{2}}$ $I_{1}=\frac{V_{1}}{\rho \frac{l_{1}}{A}}$ $I_{2}=\frac{V_{2}}{\rho \frac{l_{2}}{A}}$ From equation (i), (ii) and (iii)- $\frac{\mathrm{V}_{1}}{l_{1}} \times \frac{l_{2}}{\mathrm{~V}_{2}}=2, \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{l_{2}}{l_{1}}=2, \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{1}{2}=2$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$
153394
A current of $1 \mathrm{~A}$ is flowing along the sides of an equilateral triangle of side $4.5 \times 10^{-2} \mathrm{~m}$. The magnetic field at the centroid of the triangle is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)$
1 $4 \times 10^{-5} \mathrm{~T}$
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-4} \mathrm{~T}$
4 $2 \times 10^{-4} \mathrm{~T}$
Explanation:
A Given, $l=4.5 \times 10^{-2} \mathrm{~m}, \mathrm{I}=1 \mathrm{~A}$ Total magnetic field at the centroid due to all sides of a triangle is given by: $\mathrm{B}=\frac{3 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ For an equilateral triangle \(\theta_1=\theta_2=60^{\circ}\) From $\triangle \mathrm{BOD}$, $\tan 60^{\circ}=\mathrm{BD} / \mathrm{OD}$ $\mathrm{OD}=\mathrm{a}=\mathrm{BD} / \tan 60^{\circ}=\frac{l / 2}{\sqrt{3}} \quad(\because \mathrm{BD}=l / 2)$ $\mathrm{a}=\frac{l}{2 \sqrt{3}}=\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}$ $\therefore \quad \mathrm{B}=\frac{3 \times 4 \pi \times 10^{-7} \times 1 \times 2 \sqrt{3}}{4 \pi \times 4.5 \times 10^{-2}}\left(\frac{2 \sqrt{3}}{2}\right)$ $\mathrm{B}=4 \times 10^{-5} \mathrm{~T}$
TS EAMCET (Engg.)-2016
Moving Charges & Magnetism
153395
A square loop is made by a uniform conductor wire as shown in figure The net magnetic field at the centre of the loop if side length of the square is a
1 $\frac{\mu_{0} \mathrm{i}}{2 \mathrm{a}}$
2 Zero
3 $\frac{\mu_{0} i^{2}}{a^{2}}$
4 None of these
Explanation:
B From the figure, we can see that both part PS-R and P-Q-R have the same length so, current in both branches will be same and using the Biot-Savart law the field at the center is zero.
JIPMER-2015
Moving Charges & Magnetism
153397
A long straight wire of radius $R$ carries a steady current $I$. The current is uniformly distributed across its cross-section. The ratio of magnetic field at $R / 2$ and $2 R$ is
1 $\frac{1}{2}$
2 2
3 $\frac{1}{4}$
4 1
Explanation:
D Consider two amperion loops of radius R/2 Ans: d : Consider two amperion loops of radius $R / 2$ and 2R. Applying ampere circular law for their loops we get- For a point inside the long straight wire $\mathrm{B}_{\text {inside }}=\frac{\mu_{0} \mathrm{ir}}{2 \pi \mathrm{a}^{2}}$ $\text { at } \mathrm{r}=\frac{\mathrm{a}}{2}$ $\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ For a point out side the solid cylinder $\mathrm{B}_{\text {out }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\text { at } \mathrm{r}=2 \mathrm{a}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}} \times \frac{4 \pi \mathrm{a}}{\mu_{0} \mathrm{i}}=1$
AMU-2016
Moving Charges & Magnetism
153398
Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What potential difference ratio should be applied across them so that the magnetic field at their centers is the same?
1 2
2 3
3 4
4 6
Explanation:
C Magnetic field due to $1^{\text {st }}$ coil and second coil same at center. Then, magnetic field $\mathrm{B}$ at center is, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}_{1}}{2(2 \mathrm{r})}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \mathrm{r}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2$ As we know that, resistance of coil is related as $\mathrm{R}=\rho \frac{1}{\mathrm{~A}} \quad \text { where, } \rho =\text { resistivity }$ $1 =\text { Length }$ $\mathrm{A} =\text { Area of cross }$ $\rho$ and $A$ is same for both coil but $I_{1}=2 \pi(2 r)$ $\text { And } \quad 1_{2}=2 \pi(\mathrm{r})$ If $V_{1}$, and $V_{2}$ applied voltage first and second coil then, $I_{1}=\frac{V_{1}}{R_{1}} \text { and } I_{2}=\frac{V_{2}}{R_{2}}$ $I_{1}=\frac{V_{1}}{\rho \frac{l_{1}}{A}}$ $I_{2}=\frac{V_{2}}{\rho \frac{l_{2}}{A}}$ From equation (i), (ii) and (iii)- $\frac{\mathrm{V}_{1}}{l_{1}} \times \frac{l_{2}}{\mathrm{~V}_{2}}=2, \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{l_{2}}{l_{1}}=2, \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{1}{2}=2$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$
153394
A current of $1 \mathrm{~A}$ is flowing along the sides of an equilateral triangle of side $4.5 \times 10^{-2} \mathrm{~m}$. The magnetic field at the centroid of the triangle is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)$
1 $4 \times 10^{-5} \mathrm{~T}$
2 $2 \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-4} \mathrm{~T}$
4 $2 \times 10^{-4} \mathrm{~T}$
Explanation:
A Given, $l=4.5 \times 10^{-2} \mathrm{~m}, \mathrm{I}=1 \mathrm{~A}$ Total magnetic field at the centroid due to all sides of a triangle is given by: $\mathrm{B}=\frac{3 \mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ For an equilateral triangle \(\theta_1=\theta_2=60^{\circ}\) From $\triangle \mathrm{BOD}$, $\tan 60^{\circ}=\mathrm{BD} / \mathrm{OD}$ $\mathrm{OD}=\mathrm{a}=\mathrm{BD} / \tan 60^{\circ}=\frac{l / 2}{\sqrt{3}} \quad(\because \mathrm{BD}=l / 2)$ $\mathrm{a}=\frac{l}{2 \sqrt{3}}=\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}$ $\therefore \quad \mathrm{B}=\frac{3 \times 4 \pi \times 10^{-7} \times 1 \times 2 \sqrt{3}}{4 \pi \times 4.5 \times 10^{-2}}\left(\frac{2 \sqrt{3}}{2}\right)$ $\mathrm{B}=4 \times 10^{-5} \mathrm{~T}$
TS EAMCET (Engg.)-2016
Moving Charges & Magnetism
153395
A square loop is made by a uniform conductor wire as shown in figure The net magnetic field at the centre of the loop if side length of the square is a
1 $\frac{\mu_{0} \mathrm{i}}{2 \mathrm{a}}$
2 Zero
3 $\frac{\mu_{0} i^{2}}{a^{2}}$
4 None of these
Explanation:
B From the figure, we can see that both part PS-R and P-Q-R have the same length so, current in both branches will be same and using the Biot-Savart law the field at the center is zero.
JIPMER-2015
Moving Charges & Magnetism
153397
A long straight wire of radius $R$ carries a steady current $I$. The current is uniformly distributed across its cross-section. The ratio of magnetic field at $R / 2$ and $2 R$ is
1 $\frac{1}{2}$
2 2
3 $\frac{1}{4}$
4 1
Explanation:
D Consider two amperion loops of radius R/2 Ans: d : Consider two amperion loops of radius $R / 2$ and 2R. Applying ampere circular law for their loops we get- For a point inside the long straight wire $\mathrm{B}_{\text {inside }}=\frac{\mu_{0} \mathrm{ir}}{2 \pi \mathrm{a}^{2}}$ $\text { at } \mathrm{r}=\frac{\mathrm{a}}{2}$ $\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ For a point out side the solid cylinder $\mathrm{B}_{\text {out }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\text { at } \mathrm{r}=2 \mathrm{a}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a}} \times \frac{4 \pi \mathrm{a}}{\mu_{0} \mathrm{i}}=1$
AMU-2016
Moving Charges & Magnetism
153398
Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What potential difference ratio should be applied across them so that the magnetic field at their centers is the same?
1 2
2 3
3 4
4 6
Explanation:
C Magnetic field due to $1^{\text {st }}$ coil and second coil same at center. Then, magnetic field $\mathrm{B}$ at center is, $\mathrm{B}=\frac{\mu_{0} \mathrm{I}_{1}}{2(2 \mathrm{r})}=\frac{\mu_{0} \mathrm{I}_{2}}{2 \mathrm{r}}=\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=2$ As we know that, resistance of coil is related as $\mathrm{R}=\rho \frac{1}{\mathrm{~A}} \quad \text { where, } \rho =\text { resistivity }$ $1 =\text { Length }$ $\mathrm{A} =\text { Area of cross }$ $\rho$ and $A$ is same for both coil but $I_{1}=2 \pi(2 r)$ $\text { And } \quad 1_{2}=2 \pi(\mathrm{r})$ If $V_{1}$, and $V_{2}$ applied voltage first and second coil then, $I_{1}=\frac{V_{1}}{R_{1}} \text { and } I_{2}=\frac{V_{2}}{R_{2}}$ $I_{1}=\frac{V_{1}}{\rho \frac{l_{1}}{A}}$ $I_{2}=\frac{V_{2}}{\rho \frac{l_{2}}{A}}$ From equation (i), (ii) and (iii)- $\frac{\mathrm{V}_{1}}{l_{1}} \times \frac{l_{2}}{\mathrm{~V}_{2}}=2, \quad \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{l_{2}}{l_{1}}=2, \frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}} \times \frac{1}{2}=2$ $\mathrm{~V}_{1}=4 \mathrm{~V}_{2}$
