153910
Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is
1 $\pi$
2 $\frac{\pi}{4}$
3 $2 \pi$
4 $\frac{\pi}{2}$
Explanation:
B Let the length of the wire L Then, side o square, $\mathrm{L}=4 \mathrm{a} \Rightarrow \mathrm{a}=\mathrm{L} / 4$ Then, area $\left(\mathrm{A}_{1}\right)=(\mathrm{L} / 4)^{2}$ Radius of circular loop, $2 \pi r=L$ $r=\frac{L}{2 \pi}$ Then, area $\left(\mathrm{A}_{2}\right)=2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ We know, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\left(\mathrm{M}_{1}\right)_{\text {square }}=\mathrm{IA}_{1}$ $=\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}$ And $\quad\left(\mathrm{M}_{2}\right)_{\text {circular loop }}=\mathrm{IA}_{2}$ $=\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ From equation (i) and equation (ii), we get - $\frac{\left(M_{1}\right)_{\text {square }}}{\left(M_{2}\right)_{\text {circular loop }}}=\frac{I \times\left(\frac{L}{4}\right)^{2}}{I \times 2 \pi\left(\frac{L}{2 \pi}\right)^{2}}=\frac{\pi}{4}$
MHT-CET 2020
Moving Charges & Magnetism
153912
What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?
A Magnetic moment $(\mathrm{M})=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{R}^{2}\right)$ Magnetic field of a circular coil, $\mathrm{B}=\frac{\mu_{0} I}{2 R}$ $I=\frac{2 B R}{\mu_{0}}$ From equation (i) and (ii) $M=\frac{2 B R}{\mu_{0}} \times \pi R^{2}$ $M=\frac{2 \pi B R^{3}}{\mu_{0}}$
MHT-CET 2020
Moving Charges & Magnetism
153913
A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be
1 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
2 $2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
3 $\frac{2 \pi \mathrm{M}}{\mathrm{I}}$
4 $\frac{\pi \mathrm{M}}{\mathrm{I}}$
Explanation:
B $\mathrm{M}=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{r}^{2}\right)$ When wire of length $\mathrm{L}$ is turned into loop radius- or $\quad r=\frac{L}{2 \pi}$ Putting the value of ' $r$ ' in equation (i), we get- $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
MHT-CET 2020
Moving Charges & Magnetism
153914
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$, if the current and the radius both are doubled. The new ratio will become :
1 $2 \mathrm{x}$
2 $4 \mathrm{x}$
3 $\frac{x}{4}$
4 $\frac{x}{8}$
Explanation:
D Magnetic field of the centre of a circular loop of radius $\mathrm{R}$ carrying current $\mathrm{I}$. $\mathrm{B}=\frac{\mu_{\mathrm{o}} \cdot 2 \pi \mathrm{I}}{4 \pi \mathrm{R}}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{M}=\mathrm{I}\left(\pi \cdot \mathrm{R}^{2}\right) \quad\left(\because \mathrm{A}=\pi \mathrm{R}^{2}\right)$ $\frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_{\mathrm{o}} \cdot \mathrm{I}}{2 \mathrm{R}} \times \frac{1}{\pi \mathrm{R}^{2} \mathrm{I}}=\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}=\mathrm{x}$ When both the current and radius are doubled the ratio becomes. $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mu_{\mathrm{o}}}{2 \pi(2 \mathrm{R})^{3}}$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{1}{8}\left(\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}\right) \quad\left(\because \mathrm{x}=\frac{\mu_{0}}{2 \pi \mathrm{R}^{3}}\right)$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mathrm{x}}{8}$
153910
Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is
1 $\pi$
2 $\frac{\pi}{4}$
3 $2 \pi$
4 $\frac{\pi}{2}$
Explanation:
B Let the length of the wire L Then, side o square, $\mathrm{L}=4 \mathrm{a} \Rightarrow \mathrm{a}=\mathrm{L} / 4$ Then, area $\left(\mathrm{A}_{1}\right)=(\mathrm{L} / 4)^{2}$ Radius of circular loop, $2 \pi r=L$ $r=\frac{L}{2 \pi}$ Then, area $\left(\mathrm{A}_{2}\right)=2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ We know, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\left(\mathrm{M}_{1}\right)_{\text {square }}=\mathrm{IA}_{1}$ $=\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}$ And $\quad\left(\mathrm{M}_{2}\right)_{\text {circular loop }}=\mathrm{IA}_{2}$ $=\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ From equation (i) and equation (ii), we get - $\frac{\left(M_{1}\right)_{\text {square }}}{\left(M_{2}\right)_{\text {circular loop }}}=\frac{I \times\left(\frac{L}{4}\right)^{2}}{I \times 2 \pi\left(\frac{L}{2 \pi}\right)^{2}}=\frac{\pi}{4}$
MHT-CET 2020
Moving Charges & Magnetism
153912
What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?
A Magnetic moment $(\mathrm{M})=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{R}^{2}\right)$ Magnetic field of a circular coil, $\mathrm{B}=\frac{\mu_{0} I}{2 R}$ $I=\frac{2 B R}{\mu_{0}}$ From equation (i) and (ii) $M=\frac{2 B R}{\mu_{0}} \times \pi R^{2}$ $M=\frac{2 \pi B R^{3}}{\mu_{0}}$
MHT-CET 2020
Moving Charges & Magnetism
153913
A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be
1 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
2 $2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
3 $\frac{2 \pi \mathrm{M}}{\mathrm{I}}$
4 $\frac{\pi \mathrm{M}}{\mathrm{I}}$
Explanation:
B $\mathrm{M}=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{r}^{2}\right)$ When wire of length $\mathrm{L}$ is turned into loop radius- or $\quad r=\frac{L}{2 \pi}$ Putting the value of ' $r$ ' in equation (i), we get- $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
MHT-CET 2020
Moving Charges & Magnetism
153914
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$, if the current and the radius both are doubled. The new ratio will become :
1 $2 \mathrm{x}$
2 $4 \mathrm{x}$
3 $\frac{x}{4}$
4 $\frac{x}{8}$
Explanation:
D Magnetic field of the centre of a circular loop of radius $\mathrm{R}$ carrying current $\mathrm{I}$. $\mathrm{B}=\frac{\mu_{\mathrm{o}} \cdot 2 \pi \mathrm{I}}{4 \pi \mathrm{R}}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{M}=\mathrm{I}\left(\pi \cdot \mathrm{R}^{2}\right) \quad\left(\because \mathrm{A}=\pi \mathrm{R}^{2}\right)$ $\frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_{\mathrm{o}} \cdot \mathrm{I}}{2 \mathrm{R}} \times \frac{1}{\pi \mathrm{R}^{2} \mathrm{I}}=\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}=\mathrm{x}$ When both the current and radius are doubled the ratio becomes. $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mu_{\mathrm{o}}}{2 \pi(2 \mathrm{R})^{3}}$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{1}{8}\left(\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}\right) \quad\left(\because \mathrm{x}=\frac{\mu_{0}}{2 \pi \mathrm{R}^{3}}\right)$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mathrm{x}}{8}$
153910
Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is
1 $\pi$
2 $\frac{\pi}{4}$
3 $2 \pi$
4 $\frac{\pi}{2}$
Explanation:
B Let the length of the wire L Then, side o square, $\mathrm{L}=4 \mathrm{a} \Rightarrow \mathrm{a}=\mathrm{L} / 4$ Then, area $\left(\mathrm{A}_{1}\right)=(\mathrm{L} / 4)^{2}$ Radius of circular loop, $2 \pi r=L$ $r=\frac{L}{2 \pi}$ Then, area $\left(\mathrm{A}_{2}\right)=2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ We know, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\left(\mathrm{M}_{1}\right)_{\text {square }}=\mathrm{IA}_{1}$ $=\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}$ And $\quad\left(\mathrm{M}_{2}\right)_{\text {circular loop }}=\mathrm{IA}_{2}$ $=\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ From equation (i) and equation (ii), we get - $\frac{\left(M_{1}\right)_{\text {square }}}{\left(M_{2}\right)_{\text {circular loop }}}=\frac{I \times\left(\frac{L}{4}\right)^{2}}{I \times 2 \pi\left(\frac{L}{2 \pi}\right)^{2}}=\frac{\pi}{4}$
MHT-CET 2020
Moving Charges & Magnetism
153912
What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?
A Magnetic moment $(\mathrm{M})=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{R}^{2}\right)$ Magnetic field of a circular coil, $\mathrm{B}=\frac{\mu_{0} I}{2 R}$ $I=\frac{2 B R}{\mu_{0}}$ From equation (i) and (ii) $M=\frac{2 B R}{\mu_{0}} \times \pi R^{2}$ $M=\frac{2 \pi B R^{3}}{\mu_{0}}$
MHT-CET 2020
Moving Charges & Magnetism
153913
A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be
1 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
2 $2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
3 $\frac{2 \pi \mathrm{M}}{\mathrm{I}}$
4 $\frac{\pi \mathrm{M}}{\mathrm{I}}$
Explanation:
B $\mathrm{M}=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{r}^{2}\right)$ When wire of length $\mathrm{L}$ is turned into loop radius- or $\quad r=\frac{L}{2 \pi}$ Putting the value of ' $r$ ' in equation (i), we get- $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
MHT-CET 2020
Moving Charges & Magnetism
153914
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$, if the current and the radius both are doubled. The new ratio will become :
1 $2 \mathrm{x}$
2 $4 \mathrm{x}$
3 $\frac{x}{4}$
4 $\frac{x}{8}$
Explanation:
D Magnetic field of the centre of a circular loop of radius $\mathrm{R}$ carrying current $\mathrm{I}$. $\mathrm{B}=\frac{\mu_{\mathrm{o}} \cdot 2 \pi \mathrm{I}}{4 \pi \mathrm{R}}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{M}=\mathrm{I}\left(\pi \cdot \mathrm{R}^{2}\right) \quad\left(\because \mathrm{A}=\pi \mathrm{R}^{2}\right)$ $\frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_{\mathrm{o}} \cdot \mathrm{I}}{2 \mathrm{R}} \times \frac{1}{\pi \mathrm{R}^{2} \mathrm{I}}=\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}=\mathrm{x}$ When both the current and radius are doubled the ratio becomes. $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mu_{\mathrm{o}}}{2 \pi(2 \mathrm{R})^{3}}$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{1}{8}\left(\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}\right) \quad\left(\because \mathrm{x}=\frac{\mu_{0}}{2 \pi \mathrm{R}^{3}}\right)$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mathrm{x}}{8}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153910
Two wires of same length and material are used to form a square loop and a circular loop respectively. If same current is passed through both loops then the ratio of magnetic moment of square loop to that of circular loop is
1 $\pi$
2 $\frac{\pi}{4}$
3 $2 \pi$
4 $\frac{\pi}{2}$
Explanation:
B Let the length of the wire L Then, side o square, $\mathrm{L}=4 \mathrm{a} \Rightarrow \mathrm{a}=\mathrm{L} / 4$ Then, area $\left(\mathrm{A}_{1}\right)=(\mathrm{L} / 4)^{2}$ Radius of circular loop, $2 \pi r=L$ $r=\frac{L}{2 \pi}$ Then, area $\left(\mathrm{A}_{2}\right)=2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ We know, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\left(\mathrm{M}_{1}\right)_{\text {square }}=\mathrm{IA}_{1}$ $=\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}$ And $\quad\left(\mathrm{M}_{2}\right)_{\text {circular loop }}=\mathrm{IA}_{2}$ $=\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ From equation (i) and equation (ii), we get - $\frac{\left(M_{1}\right)_{\text {square }}}{\left(M_{2}\right)_{\text {circular loop }}}=\frac{I \times\left(\frac{L}{4}\right)^{2}}{I \times 2 \pi\left(\frac{L}{2 \pi}\right)^{2}}=\frac{\pi}{4}$
MHT-CET 2020
Moving Charges & Magnetism
153912
What is the magnetic moment of a current carrying circular coil if the radius of the circular coil is ' $R$ ' and magnetic induction at the center is ' $B$ '?
A Magnetic moment $(\mathrm{M})=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{R}^{2}\right)$ Magnetic field of a circular coil, $\mathrm{B}=\frac{\mu_{0} I}{2 R}$ $I=\frac{2 B R}{\mu_{0}}$ From equation (i) and (ii) $M=\frac{2 B R}{\mu_{0}} \times \pi R^{2}$ $M=\frac{2 \pi B R^{3}}{\mu_{0}}$
MHT-CET 2020
Moving Charges & Magnetism
153913
A straight wire carrying a current ' $I$ ' is turned into a circular loop. If the magnetic moment associated with it is ' $M$ ', the length of the wire will be
1 $\sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
2 $2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
3 $\frac{2 \pi \mathrm{M}}{\mathrm{I}}$
4 $\frac{\pi \mathrm{M}}{\mathrm{I}}$
Explanation:
B $\mathrm{M}=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{r}^{2}\right)$ When wire of length $\mathrm{L}$ is turned into loop radius- or $\quad r=\frac{L}{2 \pi}$ Putting the value of ' $r$ ' in equation (i), we get- $\mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}$ $\mathrm{~L}^{2}=\frac{4 \pi \mathrm{M}}{\mathrm{I}}$ $\mathrm{L}=2 \sqrt{\frac{\pi \mathrm{M}}{\mathrm{I}}}$
MHT-CET 2020
Moving Charges & Magnetism
153914
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$, if the current and the radius both are doubled. The new ratio will become :
1 $2 \mathrm{x}$
2 $4 \mathrm{x}$
3 $\frac{x}{4}$
4 $\frac{x}{8}$
Explanation:
D Magnetic field of the centre of a circular loop of radius $\mathrm{R}$ carrying current $\mathrm{I}$. $\mathrm{B}=\frac{\mu_{\mathrm{o}} \cdot 2 \pi \mathrm{I}}{4 \pi \mathrm{R}}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{M}=\mathrm{I}\left(\pi \cdot \mathrm{R}^{2}\right) \quad\left(\because \mathrm{A}=\pi \mathrm{R}^{2}\right)$ $\frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_{\mathrm{o}} \cdot \mathrm{I}}{2 \mathrm{R}} \times \frac{1}{\pi \mathrm{R}^{2} \mathrm{I}}=\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}=\mathrm{x}$ When both the current and radius are doubled the ratio becomes. $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mu_{\mathrm{o}}}{2 \pi(2 \mathrm{R})^{3}}$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{1}{8}\left(\frac{\mu_{\mathrm{o}}}{2 \pi \mathrm{R}^{3}}\right) \quad\left(\because \mathrm{x}=\frac{\mu_{0}}{2 \pi \mathrm{R}^{3}}\right)$ $\frac{\mathrm{B}^{\prime}}{\mathrm{M}^{\prime}}=\frac{\mathrm{x}}{8}$
