NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153400
Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $I$ and 2I, respectively. The resultant magnetic field induction at the centre will be
1 $\frac{\sqrt{5} \mu_{0} I}{2 R}$
2 $\frac{3 \mu_{0} I}{2 R}$
3 $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$
4 $\frac{\mu_{0} I}{R}$
Explanation:
A Magnetic field induction due to vertical loop at the $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ It acts in horizontal direction magnetic field induction due to horizontal loop at the center $\mathrm{O}$ is $\mathrm{B}_{2}=\frac{\mu_{0} 2 \mathrm{I}}{2 \mathrm{R}}$ It acts in vertically upward direction as B1 and B2 are perpendicular to each other. Therefore the resultant magnetic field induction of the center O $B_{\text {net }}=\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{\left(\frac{\mu_{0} I}{2 R}\right)^{2}+\left(\frac{\mu_{0} 2 I}{2 R}\right)^{2}}$ $B_{\text {net }}=\frac{\mu_{0} I}{2 R} \sqrt{1^{2}+2^{2}}, B_{\text {net }}=\frac{\sqrt{5} \mu_{0} I}{2 R}$
AIPMT-2012
Moving Charges & Magnetism
153402
A coil having $n$ number of turns and area $A$ is placed in a magnetic field $B$ so that is axis makes an angle $60^{\circ}$ with the direction of $B$. if $B$ changes with time, the magnitude of the emf induced in the coil will be
153403
A wire loop of area $A$ is placed in a uniform a magnetic field $B$ so that the direction of $B$ is parallel to the plane of the coil. If $B$ changes with time the emf induced in the loop will be
C $\overrightarrow{\mathrm{B}}$ is perpendicular to $\overrightarrow{\mathrm{A}}$ $\overrightarrow{\mathrm{B}} \perp \overrightarrow{\mathrm{A}}$ $\text { Flux, } \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}$ $\phi=\mathrm{B} A \cos 90^{\circ}$ $\phi=0$ Induced emf, $\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\mathrm{e}=0$ Hence, option (c) is correct.
WB JEE-2007
Moving Charges & Magnetism
153404
A circular current carrying coil has a radius $R$ The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $\frac{1}{8}$ th of its value at the centre of coil is
1 $\sqrt{3} \mathrm{R}$
2 $\frac{\mathrm{R}}{\sqrt{3}}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Given, $\mathrm{B}=\mathrm{B}_{\text {centre }} \times \frac{1}{8}$ The magnetic field on the axis of a current I carrying coil of radius $\mathrm{R}$ and at a distance $\mathrm{z}$ from the centre of the coil is, $\quad B=\frac{\mu_{o}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}$ $\Rightarrow \quad \frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}=\frac{1}{8}\left[\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}}{\mathrm{R}}\right]$ $\text { This gives }\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}=8 \mathrm{R}^{3}$ $\text { Solving we get, }$ $\mathrm{z}=\sqrt{3} \mathrm{R}$ Moving Charges \& Magnetism
153400
Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $I$ and 2I, respectively. The resultant magnetic field induction at the centre will be
1 $\frac{\sqrt{5} \mu_{0} I}{2 R}$
2 $\frac{3 \mu_{0} I}{2 R}$
3 $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$
4 $\frac{\mu_{0} I}{R}$
Explanation:
A Magnetic field induction due to vertical loop at the $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ It acts in horizontal direction magnetic field induction due to horizontal loop at the center $\mathrm{O}$ is $\mathrm{B}_{2}=\frac{\mu_{0} 2 \mathrm{I}}{2 \mathrm{R}}$ It acts in vertically upward direction as B1 and B2 are perpendicular to each other. Therefore the resultant magnetic field induction of the center O $B_{\text {net }}=\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{\left(\frac{\mu_{0} I}{2 R}\right)^{2}+\left(\frac{\mu_{0} 2 I}{2 R}\right)^{2}}$ $B_{\text {net }}=\frac{\mu_{0} I}{2 R} \sqrt{1^{2}+2^{2}}, B_{\text {net }}=\frac{\sqrt{5} \mu_{0} I}{2 R}$
AIPMT-2012
Moving Charges & Magnetism
153402
A coil having $n$ number of turns and area $A$ is placed in a magnetic field $B$ so that is axis makes an angle $60^{\circ}$ with the direction of $B$. if $B$ changes with time, the magnitude of the emf induced in the coil will be
153403
A wire loop of area $A$ is placed in a uniform a magnetic field $B$ so that the direction of $B$ is parallel to the plane of the coil. If $B$ changes with time the emf induced in the loop will be
C $\overrightarrow{\mathrm{B}}$ is perpendicular to $\overrightarrow{\mathrm{A}}$ $\overrightarrow{\mathrm{B}} \perp \overrightarrow{\mathrm{A}}$ $\text { Flux, } \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}$ $\phi=\mathrm{B} A \cos 90^{\circ}$ $\phi=0$ Induced emf, $\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\mathrm{e}=0$ Hence, option (c) is correct.
WB JEE-2007
Moving Charges & Magnetism
153404
A circular current carrying coil has a radius $R$ The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $\frac{1}{8}$ th of its value at the centre of coil is
1 $\sqrt{3} \mathrm{R}$
2 $\frac{\mathrm{R}}{\sqrt{3}}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Given, $\mathrm{B}=\mathrm{B}_{\text {centre }} \times \frac{1}{8}$ The magnetic field on the axis of a current I carrying coil of radius $\mathrm{R}$ and at a distance $\mathrm{z}$ from the centre of the coil is, $\quad B=\frac{\mu_{o}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}$ $\Rightarrow \quad \frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}=\frac{1}{8}\left[\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}}{\mathrm{R}}\right]$ $\text { This gives }\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}=8 \mathrm{R}^{3}$ $\text { Solving we get, }$ $\mathrm{z}=\sqrt{3} \mathrm{R}$ Moving Charges \& Magnetism
153400
Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $I$ and 2I, respectively. The resultant magnetic field induction at the centre will be
1 $\frac{\sqrt{5} \mu_{0} I}{2 R}$
2 $\frac{3 \mu_{0} I}{2 R}$
3 $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$
4 $\frac{\mu_{0} I}{R}$
Explanation:
A Magnetic field induction due to vertical loop at the $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ It acts in horizontal direction magnetic field induction due to horizontal loop at the center $\mathrm{O}$ is $\mathrm{B}_{2}=\frac{\mu_{0} 2 \mathrm{I}}{2 \mathrm{R}}$ It acts in vertically upward direction as B1 and B2 are perpendicular to each other. Therefore the resultant magnetic field induction of the center O $B_{\text {net }}=\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{\left(\frac{\mu_{0} I}{2 R}\right)^{2}+\left(\frac{\mu_{0} 2 I}{2 R}\right)^{2}}$ $B_{\text {net }}=\frac{\mu_{0} I}{2 R} \sqrt{1^{2}+2^{2}}, B_{\text {net }}=\frac{\sqrt{5} \mu_{0} I}{2 R}$
AIPMT-2012
Moving Charges & Magnetism
153402
A coil having $n$ number of turns and area $A$ is placed in a magnetic field $B$ so that is axis makes an angle $60^{\circ}$ with the direction of $B$. if $B$ changes with time, the magnitude of the emf induced in the coil will be
153403
A wire loop of area $A$ is placed in a uniform a magnetic field $B$ so that the direction of $B$ is parallel to the plane of the coil. If $B$ changes with time the emf induced in the loop will be
C $\overrightarrow{\mathrm{B}}$ is perpendicular to $\overrightarrow{\mathrm{A}}$ $\overrightarrow{\mathrm{B}} \perp \overrightarrow{\mathrm{A}}$ $\text { Flux, } \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}$ $\phi=\mathrm{B} A \cos 90^{\circ}$ $\phi=0$ Induced emf, $\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\mathrm{e}=0$ Hence, option (c) is correct.
WB JEE-2007
Moving Charges & Magnetism
153404
A circular current carrying coil has a radius $R$ The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $\frac{1}{8}$ th of its value at the centre of coil is
1 $\sqrt{3} \mathrm{R}$
2 $\frac{\mathrm{R}}{\sqrt{3}}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Given, $\mathrm{B}=\mathrm{B}_{\text {centre }} \times \frac{1}{8}$ The magnetic field on the axis of a current I carrying coil of radius $\mathrm{R}$ and at a distance $\mathrm{z}$ from the centre of the coil is, $\quad B=\frac{\mu_{o}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}$ $\Rightarrow \quad \frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}=\frac{1}{8}\left[\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}}{\mathrm{R}}\right]$ $\text { This gives }\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}=8 \mathrm{R}^{3}$ $\text { Solving we get, }$ $\mathrm{z}=\sqrt{3} \mathrm{R}$ Moving Charges \& Magnetism
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153400
Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $I$ and 2I, respectively. The resultant magnetic field induction at the centre will be
1 $\frac{\sqrt{5} \mu_{0} I}{2 R}$
2 $\frac{3 \mu_{0} I}{2 R}$
3 $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$
4 $\frac{\mu_{0} I}{R}$
Explanation:
A Magnetic field induction due to vertical loop at the $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ It acts in horizontal direction magnetic field induction due to horizontal loop at the center $\mathrm{O}$ is $\mathrm{B}_{2}=\frac{\mu_{0} 2 \mathrm{I}}{2 \mathrm{R}}$ It acts in vertically upward direction as B1 and B2 are perpendicular to each other. Therefore the resultant magnetic field induction of the center O $B_{\text {net }}=\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{\left(\frac{\mu_{0} I}{2 R}\right)^{2}+\left(\frac{\mu_{0} 2 I}{2 R}\right)^{2}}$ $B_{\text {net }}=\frac{\mu_{0} I}{2 R} \sqrt{1^{2}+2^{2}}, B_{\text {net }}=\frac{\sqrt{5} \mu_{0} I}{2 R}$
AIPMT-2012
Moving Charges & Magnetism
153402
A coil having $n$ number of turns and area $A$ is placed in a magnetic field $B$ so that is axis makes an angle $60^{\circ}$ with the direction of $B$. if $B$ changes with time, the magnitude of the emf induced in the coil will be
153403
A wire loop of area $A$ is placed in a uniform a magnetic field $B$ so that the direction of $B$ is parallel to the plane of the coil. If $B$ changes with time the emf induced in the loop will be
C $\overrightarrow{\mathrm{B}}$ is perpendicular to $\overrightarrow{\mathrm{A}}$ $\overrightarrow{\mathrm{B}} \perp \overrightarrow{\mathrm{A}}$ $\text { Flux, } \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}$ $\phi=\mathrm{B} A \cos 90^{\circ}$ $\phi=0$ Induced emf, $\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\mathrm{e}=0$ Hence, option (c) is correct.
WB JEE-2007
Moving Charges & Magnetism
153404
A circular current carrying coil has a radius $R$ The distance from the centre of the coil on the axis of the coil, where the magnetic induction is $\frac{1}{8}$ th of its value at the centre of coil is
1 $\sqrt{3} \mathrm{R}$
2 $\frac{\mathrm{R}}{\sqrt{3}}$
3 $\left(\frac{2}{\sqrt{3}}\right) \mathrm{R}$
4 $\frac{\mathrm{R}}{2 \sqrt{3}}$
Explanation:
A Given, $\mathrm{B}=\mathrm{B}_{\text {centre }} \times \frac{1}{8}$ The magnetic field on the axis of a current I carrying coil of radius $\mathrm{R}$ and at a distance $\mathrm{z}$ from the centre of the coil is, $\quad B=\frac{\mu_{o}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}$ $\Rightarrow \quad \frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}^{2}}{\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}}=\frac{1}{8}\left[\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \pi \mathrm{I}}{\mathrm{R}}\right]$ $\text { This gives }\left(\mathrm{R}^{2}+\mathrm{z}^{2}\right)^{3 / 2}=8 \mathrm{R}^{3}$ $\text { Solving we get, }$ $\mathrm{z}=\sqrt{3} \mathrm{R}$ Moving Charges \& Magnetism
