NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII04:MOVING CHARGES AND MAGNETISM
362639
An element \(\Delta \vec{l}=\Delta x \hat{i}\) is placed at the origin and carries a large current \(I = 10\;A.\) The magnetic field on the \(y\) -axis at a distance of \(0.5\,m\) from the elements \(\Delta \,x\) of \(1\,cm\) length is
1 \(12 \times {10^{ - 8}}\;T\)
2 \(4 \times {10^{ - 8}}\;T\)
3 \(10 \times {10^{ - 8}}\;T\)
4 \(8 \times {10^{ - 8}}\;T\)
Explanation:
Given that, \(\Delta \,l = \Delta \,x = l\,cm = {10^{ - 2}}\;m\) \(I = 10\,A,\,\,r = 0.5\,m\) and \(\theta = 90^\circ \) Using Biot - Savart's law, \(d B=\dfrac{\mu_{0}}{4 \pi} \dfrac{I d l \sin \theta}{r^{2}}\) or \(d B=10^{-7} \times \dfrac{10 \times 10^{-2} \times \sin 90^{\circ}}{(0.5)^{2}}\) \(\therefore \quad d B=4 \times 10^{-8} T\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362640
The magnetic induction due to an infinitely long straight wire carrying a current \(i\) at a perpendicular distance \(r\) from wire is given by
1 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{{2i}}{r}\)
2 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{r}{{2i}}\)
3 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{r}{{2i}}\)
4 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{{2i}}{r}\)
Explanation:
Magnetic lines of force due to current carrying conductor at a distance ' \(r\) ' perpendicular to the conductor is \(B=\dfrac{\mu_{o} i}{2 \pi r}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362641
Magnetic field at a distance \(r\) from an infinitely long straight conductor carrying a steady current varies as
1 \(\frac{1}{{{r^2}}}\)
2 \(\dfrac{1}{r}\)
3 \(\dfrac{1}{r^{3}}\)
4 \(\frac{1}{{\sqrt r }}\)
Explanation:
Magnetic field due to an infinitely long straight conductor carrrying steady current at a distance \(r\) from it is given by \(B = \frac{{{\mu _0}21}}{{4\pi r}}{\rm{ }}or{\rm{ }}B \propto \frac{1}{r}\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362642
Magnetic field at a point on the line of current carrying conductor is
1 Maximum
2 Infinity
3 Zero
4 Finite value
Explanation:
Biot-savart's law says 'B' due a current carrying wire \(B=\int \dfrac{\mu_{o}}{4 \pi} \dfrac{i(\overline{d l} \times \bar{r})}{r^{3}}\) On the line of current wire \(\overline{d l} \times \bar{r}=0 \quad \therefore B=0\)
362639
An element \(\Delta \vec{l}=\Delta x \hat{i}\) is placed at the origin and carries a large current \(I = 10\;A.\) The magnetic field on the \(y\) -axis at a distance of \(0.5\,m\) from the elements \(\Delta \,x\) of \(1\,cm\) length is
1 \(12 \times {10^{ - 8}}\;T\)
2 \(4 \times {10^{ - 8}}\;T\)
3 \(10 \times {10^{ - 8}}\;T\)
4 \(8 \times {10^{ - 8}}\;T\)
Explanation:
Given that, \(\Delta \,l = \Delta \,x = l\,cm = {10^{ - 2}}\;m\) \(I = 10\,A,\,\,r = 0.5\,m\) and \(\theta = 90^\circ \) Using Biot - Savart's law, \(d B=\dfrac{\mu_{0}}{4 \pi} \dfrac{I d l \sin \theta}{r^{2}}\) or \(d B=10^{-7} \times \dfrac{10 \times 10^{-2} \times \sin 90^{\circ}}{(0.5)^{2}}\) \(\therefore \quad d B=4 \times 10^{-8} T\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362640
The magnetic induction due to an infinitely long straight wire carrying a current \(i\) at a perpendicular distance \(r\) from wire is given by
1 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{{2i}}{r}\)
2 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{r}{{2i}}\)
3 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{r}{{2i}}\)
4 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{{2i}}{r}\)
Explanation:
Magnetic lines of force due to current carrying conductor at a distance ' \(r\) ' perpendicular to the conductor is \(B=\dfrac{\mu_{o} i}{2 \pi r}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362641
Magnetic field at a distance \(r\) from an infinitely long straight conductor carrying a steady current varies as
1 \(\frac{1}{{{r^2}}}\)
2 \(\dfrac{1}{r}\)
3 \(\dfrac{1}{r^{3}}\)
4 \(\frac{1}{{\sqrt r }}\)
Explanation:
Magnetic field due to an infinitely long straight conductor carrrying steady current at a distance \(r\) from it is given by \(B = \frac{{{\mu _0}21}}{{4\pi r}}{\rm{ }}or{\rm{ }}B \propto \frac{1}{r}\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362642
Magnetic field at a point on the line of current carrying conductor is
1 Maximum
2 Infinity
3 Zero
4 Finite value
Explanation:
Biot-savart's law says 'B' due a current carrying wire \(B=\int \dfrac{\mu_{o}}{4 \pi} \dfrac{i(\overline{d l} \times \bar{r})}{r^{3}}\) On the line of current wire \(\overline{d l} \times \bar{r}=0 \quad \therefore B=0\)
362639
An element \(\Delta \vec{l}=\Delta x \hat{i}\) is placed at the origin and carries a large current \(I = 10\;A.\) The magnetic field on the \(y\) -axis at a distance of \(0.5\,m\) from the elements \(\Delta \,x\) of \(1\,cm\) length is
1 \(12 \times {10^{ - 8}}\;T\)
2 \(4 \times {10^{ - 8}}\;T\)
3 \(10 \times {10^{ - 8}}\;T\)
4 \(8 \times {10^{ - 8}}\;T\)
Explanation:
Given that, \(\Delta \,l = \Delta \,x = l\,cm = {10^{ - 2}}\;m\) \(I = 10\,A,\,\,r = 0.5\,m\) and \(\theta = 90^\circ \) Using Biot - Savart's law, \(d B=\dfrac{\mu_{0}}{4 \pi} \dfrac{I d l \sin \theta}{r^{2}}\) or \(d B=10^{-7} \times \dfrac{10 \times 10^{-2} \times \sin 90^{\circ}}{(0.5)^{2}}\) \(\therefore \quad d B=4 \times 10^{-8} T\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362640
The magnetic induction due to an infinitely long straight wire carrying a current \(i\) at a perpendicular distance \(r\) from wire is given by
1 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{{2i}}{r}\)
2 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{r}{{2i}}\)
3 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{r}{{2i}}\)
4 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{{2i}}{r}\)
Explanation:
Magnetic lines of force due to current carrying conductor at a distance ' \(r\) ' perpendicular to the conductor is \(B=\dfrac{\mu_{o} i}{2 \pi r}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362641
Magnetic field at a distance \(r\) from an infinitely long straight conductor carrying a steady current varies as
1 \(\frac{1}{{{r^2}}}\)
2 \(\dfrac{1}{r}\)
3 \(\dfrac{1}{r^{3}}\)
4 \(\frac{1}{{\sqrt r }}\)
Explanation:
Magnetic field due to an infinitely long straight conductor carrrying steady current at a distance \(r\) from it is given by \(B = \frac{{{\mu _0}21}}{{4\pi r}}{\rm{ }}or{\rm{ }}B \propto \frac{1}{r}\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362642
Magnetic field at a point on the line of current carrying conductor is
1 Maximum
2 Infinity
3 Zero
4 Finite value
Explanation:
Biot-savart's law says 'B' due a current carrying wire \(B=\int \dfrac{\mu_{o}}{4 \pi} \dfrac{i(\overline{d l} \times \bar{r})}{r^{3}}\) On the line of current wire \(\overline{d l} \times \bar{r}=0 \quad \therefore B=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII04:MOVING CHARGES AND MAGNETISM
362639
An element \(\Delta \vec{l}=\Delta x \hat{i}\) is placed at the origin and carries a large current \(I = 10\;A.\) The magnetic field on the \(y\) -axis at a distance of \(0.5\,m\) from the elements \(\Delta \,x\) of \(1\,cm\) length is
1 \(12 \times {10^{ - 8}}\;T\)
2 \(4 \times {10^{ - 8}}\;T\)
3 \(10 \times {10^{ - 8}}\;T\)
4 \(8 \times {10^{ - 8}}\;T\)
Explanation:
Given that, \(\Delta \,l = \Delta \,x = l\,cm = {10^{ - 2}}\;m\) \(I = 10\,A,\,\,r = 0.5\,m\) and \(\theta = 90^\circ \) Using Biot - Savart's law, \(d B=\dfrac{\mu_{0}}{4 \pi} \dfrac{I d l \sin \theta}{r^{2}}\) or \(d B=10^{-7} \times \dfrac{10 \times 10^{-2} \times \sin 90^{\circ}}{(0.5)^{2}}\) \(\therefore \quad d B=4 \times 10^{-8} T\)
JEE - 2024
PHXII04:MOVING CHARGES AND MAGNETISM
362640
The magnetic induction due to an infinitely long straight wire carrying a current \(i\) at a perpendicular distance \(r\) from wire is given by
1 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{{2i}}{r}\)
2 \(\left| B \right| = \left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{r}{{2i}}\)
3 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{r}{{2i}}\)
4 \(\left| B \right| = \left( {\frac{{4\pi }}{{{\mu _0}}}} \right)\frac{{2i}}{r}\)
Explanation:
Magnetic lines of force due to current carrying conductor at a distance ' \(r\) ' perpendicular to the conductor is \(B=\dfrac{\mu_{o} i}{2 \pi r}\)
PHXII04:MOVING CHARGES AND MAGNETISM
362641
Magnetic field at a distance \(r\) from an infinitely long straight conductor carrying a steady current varies as
1 \(\frac{1}{{{r^2}}}\)
2 \(\dfrac{1}{r}\)
3 \(\dfrac{1}{r^{3}}\)
4 \(\frac{1}{{\sqrt r }}\)
Explanation:
Magnetic field due to an infinitely long straight conductor carrrying steady current at a distance \(r\) from it is given by \(B = \frac{{{\mu _0}21}}{{4\pi r}}{\rm{ }}or{\rm{ }}B \propto \frac{1}{r}\)
KCET - 2012
PHXII04:MOVING CHARGES AND MAGNETISM
362642
Magnetic field at a point on the line of current carrying conductor is
1 Maximum
2 Infinity
3 Zero
4 Finite value
Explanation:
Biot-savart's law says 'B' due a current carrying wire \(B=\int \dfrac{\mu_{o}}{4 \pi} \dfrac{i(\overline{d l} \times \bar{r})}{r^{3}}\) On the line of current wire \(\overline{d l} \times \bar{r}=0 \quad \therefore B=0\)
