153155
If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to
1 $\frac{1}{3}$
2 $\frac{1}{8}$
3 $\frac{1}{4}$
4 $\frac{1}{2}$
Explanation:
B Magnetic field at centre of will $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ At, $\quad r=R \sqrt{3}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}$ $B_{1}=\frac{\mu_{0} I R^{2}}{2\left(R^{2}+3 R^{2}\right)^{3 / 2}}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{16 \mathrm{R}^{3}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{8}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153158
A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$, the length of wire will be
1 $4 \pi \mathrm{IM}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{4 \pi \mathrm{I}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{4 \mathrm{I}}$
Explanation:
B Magnetic moment of straight current carrying wire- $M=I A$ $M=I\left(\pi r^{2}\right)$ Where, $r=$ radius of loop When wire of length $L$ is turned into loop of radius $r$ then $\mathrm{L}=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{\mathrm{L}}{2 \pi}$ Substituting the value of $r$ in equation (1) $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153160
A horizontal overhead power line carries a current of $90 \mathrm{~A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 \mathrm{~m}$ below the line?
1 $1.2 \times 10^{-5} \mathrm{~T}$ towards south
2 $1.2 \times 10^{-5} \mathrm{~T}$ towards north
3 $1.2 \times 10^{-5} \mathrm{~T}$ towards east
4 $1.2 \times 10^{-5} \mathrm{~T}$ towards west
Explanation:
A Magnetic field due to infinite wire, $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{i}=90 \mathrm{~A}, \mathrm{r}=1.5 \mathrm{~m}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=1.2 \times 10^{-5} \text { tesla }$ Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.
COMEDK 2020
Moving Charges & Magnetism
153162
Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 \mathrm{~cm}$ from the wire. The field at $40 \mathrm{~cm}$ from the wire would be:
1 4B
2 $\frac{B}{2}$
3 $\frac{B}{8}$
4 $\mathrm{B}$
Explanation:
C In a long straight wire, Given, $\mathrm{B}_{1}=\mathrm{B}$ $\mathrm{r}_{1}=5 \mathrm{~cm}$ $\mathrm{r}_{2}=40 \mathrm{~cm}$ $\mathrm{~B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{5}{40}$ According to the question, $\therefore \quad \frac{\mathrm{B}_{2}}{\mathrm{~B}} =\frac{1}{8}$ $\mathrm{~B}_{2} =\frac{\mathrm{B}}{8}$
153155
If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to
1 $\frac{1}{3}$
2 $\frac{1}{8}$
3 $\frac{1}{4}$
4 $\frac{1}{2}$
Explanation:
B Magnetic field at centre of will $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ At, $\quad r=R \sqrt{3}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}$ $B_{1}=\frac{\mu_{0} I R^{2}}{2\left(R^{2}+3 R^{2}\right)^{3 / 2}}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{16 \mathrm{R}^{3}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{8}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153158
A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$, the length of wire will be
1 $4 \pi \mathrm{IM}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{4 \pi \mathrm{I}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{4 \mathrm{I}}$
Explanation:
B Magnetic moment of straight current carrying wire- $M=I A$ $M=I\left(\pi r^{2}\right)$ Where, $r=$ radius of loop When wire of length $L$ is turned into loop of radius $r$ then $\mathrm{L}=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{\mathrm{L}}{2 \pi}$ Substituting the value of $r$ in equation (1) $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153160
A horizontal overhead power line carries a current of $90 \mathrm{~A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 \mathrm{~m}$ below the line?
1 $1.2 \times 10^{-5} \mathrm{~T}$ towards south
2 $1.2 \times 10^{-5} \mathrm{~T}$ towards north
3 $1.2 \times 10^{-5} \mathrm{~T}$ towards east
4 $1.2 \times 10^{-5} \mathrm{~T}$ towards west
Explanation:
A Magnetic field due to infinite wire, $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{i}=90 \mathrm{~A}, \mathrm{r}=1.5 \mathrm{~m}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=1.2 \times 10^{-5} \text { tesla }$ Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.
COMEDK 2020
Moving Charges & Magnetism
153162
Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 \mathrm{~cm}$ from the wire. The field at $40 \mathrm{~cm}$ from the wire would be:
1 4B
2 $\frac{B}{2}$
3 $\frac{B}{8}$
4 $\mathrm{B}$
Explanation:
C In a long straight wire, Given, $\mathrm{B}_{1}=\mathrm{B}$ $\mathrm{r}_{1}=5 \mathrm{~cm}$ $\mathrm{r}_{2}=40 \mathrm{~cm}$ $\mathrm{~B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{5}{40}$ According to the question, $\therefore \quad \frac{\mathrm{B}_{2}}{\mathrm{~B}} =\frac{1}{8}$ $\mathrm{~B}_{2} =\frac{\mathrm{B}}{8}$
153155
If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to
1 $\frac{1}{3}$
2 $\frac{1}{8}$
3 $\frac{1}{4}$
4 $\frac{1}{2}$
Explanation:
B Magnetic field at centre of will $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ At, $\quad r=R \sqrt{3}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}$ $B_{1}=\frac{\mu_{0} I R^{2}}{2\left(R^{2}+3 R^{2}\right)^{3 / 2}}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{16 \mathrm{R}^{3}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{8}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153158
A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$, the length of wire will be
1 $4 \pi \mathrm{IM}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{4 \pi \mathrm{I}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{4 \mathrm{I}}$
Explanation:
B Magnetic moment of straight current carrying wire- $M=I A$ $M=I\left(\pi r^{2}\right)$ Where, $r=$ radius of loop When wire of length $L$ is turned into loop of radius $r$ then $\mathrm{L}=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{\mathrm{L}}{2 \pi}$ Substituting the value of $r$ in equation (1) $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153160
A horizontal overhead power line carries a current of $90 \mathrm{~A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 \mathrm{~m}$ below the line?
1 $1.2 \times 10^{-5} \mathrm{~T}$ towards south
2 $1.2 \times 10^{-5} \mathrm{~T}$ towards north
3 $1.2 \times 10^{-5} \mathrm{~T}$ towards east
4 $1.2 \times 10^{-5} \mathrm{~T}$ towards west
Explanation:
A Magnetic field due to infinite wire, $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{i}=90 \mathrm{~A}, \mathrm{r}=1.5 \mathrm{~m}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=1.2 \times 10^{-5} \text { tesla }$ Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.
COMEDK 2020
Moving Charges & Magnetism
153162
Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 \mathrm{~cm}$ from the wire. The field at $40 \mathrm{~cm}$ from the wire would be:
1 4B
2 $\frac{B}{2}$
3 $\frac{B}{8}$
4 $\mathrm{B}$
Explanation:
C In a long straight wire, Given, $\mathrm{B}_{1}=\mathrm{B}$ $\mathrm{r}_{1}=5 \mathrm{~cm}$ $\mathrm{r}_{2}=40 \mathrm{~cm}$ $\mathrm{~B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{5}{40}$ According to the question, $\therefore \quad \frac{\mathrm{B}_{2}}{\mathrm{~B}} =\frac{1}{8}$ $\mathrm{~B}_{2} =\frac{\mathrm{B}}{8}$
153155
If $B_{1}$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_{2}$ is the magnetic field at the centre of the coil, then the ratio of $\frac{B_{1}}{B_{2}}$ is equal to
1 $\frac{1}{3}$
2 $\frac{1}{8}$
3 $\frac{1}{4}$
4 $\frac{1}{2}$
Explanation:
B Magnetic field at centre of will $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ At, $\quad r=R \sqrt{3}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)^{3 / 2}}$ $B_{1}=\frac{\mu_{0} I R^{2}}{2\left(R^{2}+3 R^{2}\right)^{3 / 2}}$ $\mathrm{B}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{IR}^{2}}{16 \mathrm{R}^{3}}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{1}{8}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153158
A straight wire carrying current $I$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is $M$, the length of wire will be
1 $4 \pi \mathrm{IM}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
3 $\sqrt{\frac{4 \pi \mathrm{I}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{4 \mathrm{I}}$
Explanation:
B Magnetic moment of straight current carrying wire- $M=I A$ $M=I\left(\pi r^{2}\right)$ Where, $r=$ radius of loop When wire of length $L$ is turned into loop of radius $r$ then $\mathrm{L}=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{\mathrm{L}}{2 \pi}$ Substituting the value of $r$ in equation (1) $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{I}}}$
AP EAMCET (18.09.2020) Shift-II
Moving Charges & Magnetism
153160
A horizontal overhead power line carries a current of $90 \mathrm{~A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 \mathrm{~m}$ below the line?
1 $1.2 \times 10^{-5} \mathrm{~T}$ towards south
2 $1.2 \times 10^{-5} \mathrm{~T}$ towards north
3 $1.2 \times 10^{-5} \mathrm{~T}$ towards east
4 $1.2 \times 10^{-5} \mathrm{~T}$ towards west
Explanation:
A Magnetic field due to infinite wire, $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \pi \mathrm{r}}$ $\mathrm{i}=90 \mathrm{~A}, \mathrm{r}=1.5 \mathrm{~m}$ $\mathrm{~B}=\frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$ $\mathrm{~B}=1.2 \times 10^{-5} \text { tesla }$ Direction of magnetic field is finding out by right hand rule or palm rule which is towards south.
COMEDK 2020
Moving Charges & Magnetism
153162
Consider a long straight conducting wire. The magnetic field is determined as $B$ at a distance of $5 \mathrm{~cm}$ from the wire. The field at $40 \mathrm{~cm}$ from the wire would be:
1 4B
2 $\frac{B}{2}$
3 $\frac{B}{8}$
4 $\mathrm{B}$
Explanation:
C In a long straight wire, Given, $\mathrm{B}_{1}=\mathrm{B}$ $\mathrm{r}_{1}=5 \mathrm{~cm}$ $\mathrm{r}_{2}=40 \mathrm{~cm}$ $\mathrm{~B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$ $\mathrm{B} \propto \frac{1}{\mathrm{r}}$ $\frac{\mathrm{B}_{2}}{\mathrm{~B}_{1}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{5}{40}$ According to the question, $\therefore \quad \frac{\mathrm{B}_{2}}{\mathrm{~B}} =\frac{1}{8}$ $\mathrm{~B}_{2} =\frac{\mathrm{B}}{8}$
