153302
Magnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as:
1 $\frac{1}{\sqrt{\mathrm{r}}}$
2 $\frac{1}{\mathrm{r}^{2}}$
3 $\frac{1}{\mathrm{r}}$
4 $\frac{1}{\mathrm{r}^{3}}$
Explanation:
C Magnetic field at a distance $r$ due to an infinitely long straight conductor, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B} \propto \frac{1}{\mathrm{r}}$
Karnataka CET-2012
Moving Charges & Magnetism
153310
The direction of magnetic field $d \vec{B}$ due to a current element $I \mathrm{~d} \vec{l}$ at a point of distance $r$ from it, when a current I passes through a long conductor is in the direction
1 of position vector $\vec{r}$ of the point
2 of current element $\mathrm{d} \vec{l}$
3 perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$
4 perpendicular to $\mathrm{d} \vec{l}$ only
Explanation:
C The direction of magnetic field $d \vec{B}$ due to a current element I. $\mathrm{d} \vec{l}$ at a point of distance $\mathrm{r}$ from it. When the current passes through a long conductor is in the direction perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$ By the Biot - savart law states. $\mathrm{dB}=\frac{\mu_{0} \mathrm{Id} l \times \mathrm{r}}{4 \pi \mathrm{r}^{3}}$ $\therefore$ Direction of $\mathrm{dB}$ is perpendicular to both $\mathrm{d} l$ and $\mathrm{r}$
J and K CET- 2008
Moving Charges & Magnetism
153313
A wire oriented in the east-west direction carries a current eastward. Direction of the magnetic field at a point to the south of wire is
1 vertically down
2 vertically up
3 north-east
4 south east
Explanation:
A According to right hand thumb rule when current eastward then direction of magnetic field at a point south of wire is vertically downward
J and K CET- 2006
Moving Charges & Magnetism
153315
Two thin long parallel wires separated by a distance $b$ are carrying current $I$ amp each. The magnitude of the force per unit length exerted by one wire on the other is
A We know that force per unit length two thin long parallel wire distance 'b' $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{b}}$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I} \mathrm{I}}{2 \pi \mathrm{b}}\left(\therefore \mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}\right)$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153302
Magnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as:
1 $\frac{1}{\sqrt{\mathrm{r}}}$
2 $\frac{1}{\mathrm{r}^{2}}$
3 $\frac{1}{\mathrm{r}}$
4 $\frac{1}{\mathrm{r}^{3}}$
Explanation:
C Magnetic field at a distance $r$ due to an infinitely long straight conductor, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B} \propto \frac{1}{\mathrm{r}}$
Karnataka CET-2012
Moving Charges & Magnetism
153310
The direction of magnetic field $d \vec{B}$ due to a current element $I \mathrm{~d} \vec{l}$ at a point of distance $r$ from it, when a current I passes through a long conductor is in the direction
1 of position vector $\vec{r}$ of the point
2 of current element $\mathrm{d} \vec{l}$
3 perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$
4 perpendicular to $\mathrm{d} \vec{l}$ only
Explanation:
C The direction of magnetic field $d \vec{B}$ due to a current element I. $\mathrm{d} \vec{l}$ at a point of distance $\mathrm{r}$ from it. When the current passes through a long conductor is in the direction perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$ By the Biot - savart law states. $\mathrm{dB}=\frac{\mu_{0} \mathrm{Id} l \times \mathrm{r}}{4 \pi \mathrm{r}^{3}}$ $\therefore$ Direction of $\mathrm{dB}$ is perpendicular to both $\mathrm{d} l$ and $\mathrm{r}$
J and K CET- 2008
Moving Charges & Magnetism
153313
A wire oriented in the east-west direction carries a current eastward. Direction of the magnetic field at a point to the south of wire is
1 vertically down
2 vertically up
3 north-east
4 south east
Explanation:
A According to right hand thumb rule when current eastward then direction of magnetic field at a point south of wire is vertically downward
J and K CET- 2006
Moving Charges & Magnetism
153315
Two thin long parallel wires separated by a distance $b$ are carrying current $I$ amp each. The magnitude of the force per unit length exerted by one wire on the other is
A We know that force per unit length two thin long parallel wire distance 'b' $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{b}}$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I} \mathrm{I}}{2 \pi \mathrm{b}}\left(\therefore \mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}\right)$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}}$
153302
Magnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as:
1 $\frac{1}{\sqrt{\mathrm{r}}}$
2 $\frac{1}{\mathrm{r}^{2}}$
3 $\frac{1}{\mathrm{r}}$
4 $\frac{1}{\mathrm{r}^{3}}$
Explanation:
C Magnetic field at a distance $r$ due to an infinitely long straight conductor, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B} \propto \frac{1}{\mathrm{r}}$
Karnataka CET-2012
Moving Charges & Magnetism
153310
The direction of magnetic field $d \vec{B}$ due to a current element $I \mathrm{~d} \vec{l}$ at a point of distance $r$ from it, when a current I passes through a long conductor is in the direction
1 of position vector $\vec{r}$ of the point
2 of current element $\mathrm{d} \vec{l}$
3 perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$
4 perpendicular to $\mathrm{d} \vec{l}$ only
Explanation:
C The direction of magnetic field $d \vec{B}$ due to a current element I. $\mathrm{d} \vec{l}$ at a point of distance $\mathrm{r}$ from it. When the current passes through a long conductor is in the direction perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$ By the Biot - savart law states. $\mathrm{dB}=\frac{\mu_{0} \mathrm{Id} l \times \mathrm{r}}{4 \pi \mathrm{r}^{3}}$ $\therefore$ Direction of $\mathrm{dB}$ is perpendicular to both $\mathrm{d} l$ and $\mathrm{r}$
J and K CET- 2008
Moving Charges & Magnetism
153313
A wire oriented in the east-west direction carries a current eastward. Direction of the magnetic field at a point to the south of wire is
1 vertically down
2 vertically up
3 north-east
4 south east
Explanation:
A According to right hand thumb rule when current eastward then direction of magnetic field at a point south of wire is vertically downward
J and K CET- 2006
Moving Charges & Magnetism
153315
Two thin long parallel wires separated by a distance $b$ are carrying current $I$ amp each. The magnitude of the force per unit length exerted by one wire on the other is
A We know that force per unit length two thin long parallel wire distance 'b' $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{b}}$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I} \mathrm{I}}{2 \pi \mathrm{b}}\left(\therefore \mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}\right)$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}}$
153302
Magnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as:
1 $\frac{1}{\sqrt{\mathrm{r}}}$
2 $\frac{1}{\mathrm{r}^{2}}$
3 $\frac{1}{\mathrm{r}}$
4 $\frac{1}{\mathrm{r}^{3}}$
Explanation:
C Magnetic field at a distance $r$ due to an infinitely long straight conductor, $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B} \propto \frac{1}{\mathrm{r}}$
Karnataka CET-2012
Moving Charges & Magnetism
153310
The direction of magnetic field $d \vec{B}$ due to a current element $I \mathrm{~d} \vec{l}$ at a point of distance $r$ from it, when a current I passes through a long conductor is in the direction
1 of position vector $\vec{r}$ of the point
2 of current element $\mathrm{d} \vec{l}$
3 perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$
4 perpendicular to $\mathrm{d} \vec{l}$ only
Explanation:
C The direction of magnetic field $d \vec{B}$ due to a current element I. $\mathrm{d} \vec{l}$ at a point of distance $\mathrm{r}$ from it. When the current passes through a long conductor is in the direction perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$ By the Biot - savart law states. $\mathrm{dB}=\frac{\mu_{0} \mathrm{Id} l \times \mathrm{r}}{4 \pi \mathrm{r}^{3}}$ $\therefore$ Direction of $\mathrm{dB}$ is perpendicular to both $\mathrm{d} l$ and $\mathrm{r}$
J and K CET- 2008
Moving Charges & Magnetism
153313
A wire oriented in the east-west direction carries a current eastward. Direction of the magnetic field at a point to the south of wire is
1 vertically down
2 vertically up
3 north-east
4 south east
Explanation:
A According to right hand thumb rule when current eastward then direction of magnetic field at a point south of wire is vertically downward
J and K CET- 2006
Moving Charges & Magnetism
153315
Two thin long parallel wires separated by a distance $b$ are carrying current $I$ amp each. The magnitude of the force per unit length exerted by one wire on the other is
A We know that force per unit length two thin long parallel wire distance 'b' $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{2 \pi \mathrm{b}}$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I} \mathrm{I}}{2 \pi \mathrm{b}}\left(\therefore \mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}\right)$ $\mathrm{F} / l=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi \mathrm{b}}$
