153287
Let a straight wire of length $l$ carries a current i. The magnitude of magnetic field produced by the current at point $P$ (as shown in figure) is-
B According to figure $\tan \theta=\frac{l}{l}=1$ $\theta=\frac{\pi}{4}=45^{\circ}$ Magnetic field at $\mathrm{P}$ due to straight wire of length $l$, $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin 0^{\circ}+\sin 45^{\circ}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l} \times \frac{1}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{4 \sqrt{2} \pi l}$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\sqrt{2} \mu_{0} \mathrm{i}}{8 \pi l}$
BCECE-2012
Moving Charges & Magnetism
153290
If a current $I$ is flowing in a loop of radius $r$ as shown in adjoining figure, then the magnetic field induction at the centre $O$ will be
B The current flowing in wire = I Radius of the wire $=r$ As we know magnetic field at the centre of loop, Magnetic field $(B)=\frac{\mu_{0}}{4 \pi} \times \frac{I \theta}{r}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I} \theta}{4 \pi \mathrm{r}}$
VITEEE-2013
Moving Charges & Magnetism
153291
Charge $q$ is uniformly spread on a thin ring of radius $R$. The ring rotates about its axis with a uniform frequency $f \mathrm{~Hz}$. The magnitude of magnetic induction at the centre of the ring is
1 $\frac{\mu_{0} q f}{2 R}$
2 $\frac{\mu_{0} q}{2 \mathrm{fR}}$
3 $\frac{\mu_{0} q}{2 \pi f R}$
4 $\frac{\mu_{0} \mathrm{qf}}{2 \pi \mathrm{R}}$
Explanation:
A We know that, $\mathrm{q}=$ it $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{q} \times \mathrm{f}$ $\left(\therefore \frac{1}{\mathrm{t}}=\mathrm{f}\right)$ Magnetic field at the centre of current carrying ring $B=\frac{\mu_{0} i}{2 R}$ $B=\frac{\mu_{0} q f}{2 R}$
VITEEE-2012
Moving Charges & Magnetism
153292
In hydrogen atom, an electron is revolving in the orbit of radius $0.53 \AA$ with $6.6 \times 10^{15}$ rps. Magnetic field produced at the centre of the orbit is
1 $0.125 \mathrm{~Wb} / \mathrm{m}^{2}$
2 $1.25 \mathrm{~Wb} / \mathrm{m}^{2}$
3 $12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
4 $125 \mathrm{~Wb} / \mathrm{m}^{2}$
Explanation:
C Given, $\mathrm{r}=0.53 \AA$ $\mathrm{f}=6.6 \times 10^{15} \mathrm{rps}$ We know that, Current (i) $=\mathrm{q} \times \mathrm{f}$ $\mathrm{i}=1.6 \times 10^{-19} \times 6.6 \times 10^{15}=1.056 \times 10^{-3} \mathrm{~A}$ Then magnetic field- $B=\frac{\mu_{0} I}{2 r}$ $B=\frac{4 \pi \times 10^{-7} \times 1.056 \times 10^{-3}}{2 \times 0.53 \times 10^{-10}}=12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
153287
Let a straight wire of length $l$ carries a current i. The magnitude of magnetic field produced by the current at point $P$ (as shown in figure) is-
B According to figure $\tan \theta=\frac{l}{l}=1$ $\theta=\frac{\pi}{4}=45^{\circ}$ Magnetic field at $\mathrm{P}$ due to straight wire of length $l$, $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin 0^{\circ}+\sin 45^{\circ}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l} \times \frac{1}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{4 \sqrt{2} \pi l}$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\sqrt{2} \mu_{0} \mathrm{i}}{8 \pi l}$
BCECE-2012
Moving Charges & Magnetism
153290
If a current $I$ is flowing in a loop of radius $r$ as shown in adjoining figure, then the magnetic field induction at the centre $O$ will be
B The current flowing in wire = I Radius of the wire $=r$ As we know magnetic field at the centre of loop, Magnetic field $(B)=\frac{\mu_{0}}{4 \pi} \times \frac{I \theta}{r}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I} \theta}{4 \pi \mathrm{r}}$
VITEEE-2013
Moving Charges & Magnetism
153291
Charge $q$ is uniformly spread on a thin ring of radius $R$. The ring rotates about its axis with a uniform frequency $f \mathrm{~Hz}$. The magnitude of magnetic induction at the centre of the ring is
1 $\frac{\mu_{0} q f}{2 R}$
2 $\frac{\mu_{0} q}{2 \mathrm{fR}}$
3 $\frac{\mu_{0} q}{2 \pi f R}$
4 $\frac{\mu_{0} \mathrm{qf}}{2 \pi \mathrm{R}}$
Explanation:
A We know that, $\mathrm{q}=$ it $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{q} \times \mathrm{f}$ $\left(\therefore \frac{1}{\mathrm{t}}=\mathrm{f}\right)$ Magnetic field at the centre of current carrying ring $B=\frac{\mu_{0} i}{2 R}$ $B=\frac{\mu_{0} q f}{2 R}$
VITEEE-2012
Moving Charges & Magnetism
153292
In hydrogen atom, an electron is revolving in the orbit of radius $0.53 \AA$ with $6.6 \times 10^{15}$ rps. Magnetic field produced at the centre of the orbit is
1 $0.125 \mathrm{~Wb} / \mathrm{m}^{2}$
2 $1.25 \mathrm{~Wb} / \mathrm{m}^{2}$
3 $12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
4 $125 \mathrm{~Wb} / \mathrm{m}^{2}$
Explanation:
C Given, $\mathrm{r}=0.53 \AA$ $\mathrm{f}=6.6 \times 10^{15} \mathrm{rps}$ We know that, Current (i) $=\mathrm{q} \times \mathrm{f}$ $\mathrm{i}=1.6 \times 10^{-19} \times 6.6 \times 10^{15}=1.056 \times 10^{-3} \mathrm{~A}$ Then magnetic field- $B=\frac{\mu_{0} I}{2 r}$ $B=\frac{4 \pi \times 10^{-7} \times 1.056 \times 10^{-3}}{2 \times 0.53 \times 10^{-10}}=12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
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Moving Charges & Magnetism
153287
Let a straight wire of length $l$ carries a current i. The magnitude of magnetic field produced by the current at point $P$ (as shown in figure) is-
B According to figure $\tan \theta=\frac{l}{l}=1$ $\theta=\frac{\pi}{4}=45^{\circ}$ Magnetic field at $\mathrm{P}$ due to straight wire of length $l$, $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin 0^{\circ}+\sin 45^{\circ}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l} \times \frac{1}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{4 \sqrt{2} \pi l}$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\sqrt{2} \mu_{0} \mathrm{i}}{8 \pi l}$
BCECE-2012
Moving Charges & Magnetism
153290
If a current $I$ is flowing in a loop of radius $r$ as shown in adjoining figure, then the magnetic field induction at the centre $O$ will be
B The current flowing in wire = I Radius of the wire $=r$ As we know magnetic field at the centre of loop, Magnetic field $(B)=\frac{\mu_{0}}{4 \pi} \times \frac{I \theta}{r}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I} \theta}{4 \pi \mathrm{r}}$
VITEEE-2013
Moving Charges & Magnetism
153291
Charge $q$ is uniformly spread on a thin ring of radius $R$. The ring rotates about its axis with a uniform frequency $f \mathrm{~Hz}$. The magnitude of magnetic induction at the centre of the ring is
1 $\frac{\mu_{0} q f}{2 R}$
2 $\frac{\mu_{0} q}{2 \mathrm{fR}}$
3 $\frac{\mu_{0} q}{2 \pi f R}$
4 $\frac{\mu_{0} \mathrm{qf}}{2 \pi \mathrm{R}}$
Explanation:
A We know that, $\mathrm{q}=$ it $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{q} \times \mathrm{f}$ $\left(\therefore \frac{1}{\mathrm{t}}=\mathrm{f}\right)$ Magnetic field at the centre of current carrying ring $B=\frac{\mu_{0} i}{2 R}$ $B=\frac{\mu_{0} q f}{2 R}$
VITEEE-2012
Moving Charges & Magnetism
153292
In hydrogen atom, an electron is revolving in the orbit of radius $0.53 \AA$ with $6.6 \times 10^{15}$ rps. Magnetic field produced at the centre of the orbit is
1 $0.125 \mathrm{~Wb} / \mathrm{m}^{2}$
2 $1.25 \mathrm{~Wb} / \mathrm{m}^{2}$
3 $12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
4 $125 \mathrm{~Wb} / \mathrm{m}^{2}$
Explanation:
C Given, $\mathrm{r}=0.53 \AA$ $\mathrm{f}=6.6 \times 10^{15} \mathrm{rps}$ We know that, Current (i) $=\mathrm{q} \times \mathrm{f}$ $\mathrm{i}=1.6 \times 10^{-19} \times 6.6 \times 10^{15}=1.056 \times 10^{-3} \mathrm{~A}$ Then magnetic field- $B=\frac{\mu_{0} I}{2 r}$ $B=\frac{4 \pi \times 10^{-7} \times 1.056 \times 10^{-3}}{2 \times 0.53 \times 10^{-10}}=12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
153287
Let a straight wire of length $l$ carries a current i. The magnitude of magnetic field produced by the current at point $P$ (as shown in figure) is-
B According to figure $\tan \theta=\frac{l}{l}=1$ $\theta=\frac{\pi}{4}=45^{\circ}$ Magnetic field at $\mathrm{P}$ due to straight wire of length $l$, $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin \theta_{1}+\sin \theta_{2}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l}\left(\sin 0^{\circ}+\sin 45^{\circ}\right)$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\mu_{0} \mathrm{i}}{4 \pi l} \times \frac{1}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{4 \sqrt{2} \pi l}$ $\overrightarrow{\mathrm{B}}_{\mathrm{P}}=\frac{\sqrt{2} \mu_{0} \mathrm{i}}{8 \pi l}$
BCECE-2012
Moving Charges & Magnetism
153290
If a current $I$ is flowing in a loop of radius $r$ as shown in adjoining figure, then the magnetic field induction at the centre $O$ will be
B The current flowing in wire = I Radius of the wire $=r$ As we know magnetic field at the centre of loop, Magnetic field $(B)=\frac{\mu_{0}}{4 \pi} \times \frac{I \theta}{r}$ $\mathrm{B}=\frac{\mu_{0} \mathrm{I} \theta}{4 \pi \mathrm{r}}$
VITEEE-2013
Moving Charges & Magnetism
153291
Charge $q$ is uniformly spread on a thin ring of radius $R$. The ring rotates about its axis with a uniform frequency $f \mathrm{~Hz}$. The magnitude of magnetic induction at the centre of the ring is
1 $\frac{\mu_{0} q f}{2 R}$
2 $\frac{\mu_{0} q}{2 \mathrm{fR}}$
3 $\frac{\mu_{0} q}{2 \pi f R}$
4 $\frac{\mu_{0} \mathrm{qf}}{2 \pi \mathrm{R}}$
Explanation:
A We know that, $\mathrm{q}=$ it $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{q} \times \mathrm{f}$ $\left(\therefore \frac{1}{\mathrm{t}}=\mathrm{f}\right)$ Magnetic field at the centre of current carrying ring $B=\frac{\mu_{0} i}{2 R}$ $B=\frac{\mu_{0} q f}{2 R}$
VITEEE-2012
Moving Charges & Magnetism
153292
In hydrogen atom, an electron is revolving in the orbit of radius $0.53 \AA$ with $6.6 \times 10^{15}$ rps. Magnetic field produced at the centre of the orbit is
1 $0.125 \mathrm{~Wb} / \mathrm{m}^{2}$
2 $1.25 \mathrm{~Wb} / \mathrm{m}^{2}$
3 $12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
4 $125 \mathrm{~Wb} / \mathrm{m}^{2}$
Explanation:
C Given, $\mathrm{r}=0.53 \AA$ $\mathrm{f}=6.6 \times 10^{15} \mathrm{rps}$ We know that, Current (i) $=\mathrm{q} \times \mathrm{f}$ $\mathrm{i}=1.6 \times 10^{-19} \times 6.6 \times 10^{15}=1.056 \times 10^{-3} \mathrm{~A}$ Then magnetic field- $B=\frac{\mu_{0} I}{2 r}$ $B=\frac{4 \pi \times 10^{-7} \times 1.056 \times 10^{-3}}{2 \times 0.53 \times 10^{-10}}=12.5 \mathrm{~Wb} / \mathrm{m}^{2}$
