153901
A particle of charge ' $q$ ' and mass ' $m$ ' moves in a circular orbit of radius ' $r$ ' with angular speed ' $\omega$ '. The ratio of the magnitude of its magnetic moment to that its angular momentum is
1 $\frac{q \omega r^{2}}{2 m}$
2 $\frac{q \omega r^{2}}{2}$
3 $\frac{q \omega}{2 m r^{2}}$
4 $\frac{\mathrm{q}}{2 \mathrm{~m}}$
Explanation:
D Given, radius of orbit $=r$, magnetic moment $=\mathrm{M}$, angular speed $=\theta$ $\therefore \quad \mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}} \quad$ (circumference of circle $=2 \pi \mathrm{r}$ ) And, $\quad I=\frac{q}{t}=\frac{q v}{2 \pi r}$ $\therefore$ Magnetic moment $(\mathrm{M})=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{qvr}}{2}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ Dividing equation (i) and (ii), we get- $\frac{M}{L}=\frac{\frac{q v r}{2}}{m v r}$ $\frac{M}{L}=\frac{q}{2 m}$
MHT-CET 2020
Moving Charges & Magnetism
153902
Which one of the following proportionality represents the relation between orbital magnetic moment $M_{0}$ and orbital angular momentum $L_{0}$ of an electron?
D We know that, magnetic moment (M) = IA And, $t=\frac{2 \pi r}{v}$ $\therefore \quad \mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{q}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ So, $\quad \mathrm{M}_{0}=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2} \quad\left(\pi \mathrm{r}^{2}=\right.$ area of circle $)$ $\mathrm{M}_{0}=\frac{\mathrm{evr}}{2} \quad \ldots . \text { (i) } \quad(\because \mathrm{q}=\mathrm{e})$ Angular momentum $\left(\mathrm{L}_{0}\right)=\mathrm{mvr}$ From equation (ii), we get- $\mathrm{vr}=\frac{\mathrm{L}_{0}}{\mathrm{~m}}$ Putting the value of 'vr' in equation (i), we get- $\mathrm{M}_{0}=\frac{\mathrm{e}}{2} \times \frac{\mathrm{L}_{0}}{\mathrm{~m}}$ $\therefore \quad \mathrm{M}_{0} \propto \mathrm{L}_{0}$
MHT-CET 2020
Moving Charges & Magnetism
153906
An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is $(m=$ mass of electron, $h=$ Planck's constant, $e$ $=$ electronic charge)
1 $\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
2 $\frac{-\mathrm{eh}}{\pi \mathrm{m}}$
3 $\frac{\mathrm{eh}}{2 \pi \mathrm{m}}$
4 $\frac{2 \mathrm{eh}}{\pi \mathrm{m}}$
Explanation:
A According to Bohr's theory- $\operatorname{mvR}=\frac{\mathrm{h}}{2 \pi}$ or $\quad \mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}$ We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}$ And, $\quad I=\frac{|\mathrm{e}| \overrightarrow{\mathrm{A}}}{\mathrm{T}}$ Magnetic dipole moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M}=\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}} \times \pi \mathrm{R}^{2}$ $\mathrm{M}=\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
MHT-CET 2020
Moving Charges & Magnetism
153907
An electron revolving in circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment. ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{4}$
2 $\frac{M}{2}$
3 $\mathrm{M}$
4 $2 \mathrm{M}$
Explanation:
D We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And $\quad \mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}=\mathrm{ef}$ $\left(\because \mathrm{f}=\frac{1}{\mathrm{~T}}\right)$ $\mathrm{I} \propto \mathrm{f}$ Therefore, $\mathrm{M} \propto \mathrm{I} \propto \mathrm{f}$ Then, $\frac{M_{1}}{M_{2}}=\frac{f_{1}}{f_{2}}$ $\left(\because \mathrm{M}_{1}=\mathrm{M}, \mathrm{f}_{2}=2 \mathrm{f}_{1}\right)$ $\mathrm{M}_{2}=2 \mathrm{M}$
MHT-CET 2020
Moving Charges & Magnetism
153909
A circular and a square coil is prepared from two identical metal wires and a current is passed through them. Ratio of magnetic dipole moment associated with circular coil to that with square coil is
1 $\frac{2}{\pi}$
2 $\frac{4}{\pi}$
3 $\frac{\pi}{2}$
4 $\pi$
Explanation:
B Let the length of the wire $\mathrm{L}$ 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 - $\frac{\left(\mathrm{M}_{2}\right)_{\text {circular loop }}}{\left(\mathrm{M}_{1}\right)_{\text {square }}}=\frac{\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}}{\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}}=\frac{4}{\pi}$
153901
A particle of charge ' $q$ ' and mass ' $m$ ' moves in a circular orbit of radius ' $r$ ' with angular speed ' $\omega$ '. The ratio of the magnitude of its magnetic moment to that its angular momentum is
1 $\frac{q \omega r^{2}}{2 m}$
2 $\frac{q \omega r^{2}}{2}$
3 $\frac{q \omega}{2 m r^{2}}$
4 $\frac{\mathrm{q}}{2 \mathrm{~m}}$
Explanation:
D Given, radius of orbit $=r$, magnetic moment $=\mathrm{M}$, angular speed $=\theta$ $\therefore \quad \mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}} \quad$ (circumference of circle $=2 \pi \mathrm{r}$ ) And, $\quad I=\frac{q}{t}=\frac{q v}{2 \pi r}$ $\therefore$ Magnetic moment $(\mathrm{M})=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{qvr}}{2}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ Dividing equation (i) and (ii), we get- $\frac{M}{L}=\frac{\frac{q v r}{2}}{m v r}$ $\frac{M}{L}=\frac{q}{2 m}$
MHT-CET 2020
Moving Charges & Magnetism
153902
Which one of the following proportionality represents the relation between orbital magnetic moment $M_{0}$ and orbital angular momentum $L_{0}$ of an electron?
D We know that, magnetic moment (M) = IA And, $t=\frac{2 \pi r}{v}$ $\therefore \quad \mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{q}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ So, $\quad \mathrm{M}_{0}=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2} \quad\left(\pi \mathrm{r}^{2}=\right.$ area of circle $)$ $\mathrm{M}_{0}=\frac{\mathrm{evr}}{2} \quad \ldots . \text { (i) } \quad(\because \mathrm{q}=\mathrm{e})$ Angular momentum $\left(\mathrm{L}_{0}\right)=\mathrm{mvr}$ From equation (ii), we get- $\mathrm{vr}=\frac{\mathrm{L}_{0}}{\mathrm{~m}}$ Putting the value of 'vr' in equation (i), we get- $\mathrm{M}_{0}=\frac{\mathrm{e}}{2} \times \frac{\mathrm{L}_{0}}{\mathrm{~m}}$ $\therefore \quad \mathrm{M}_{0} \propto \mathrm{L}_{0}$
MHT-CET 2020
Moving Charges & Magnetism
153906
An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is $(m=$ mass of electron, $h=$ Planck's constant, $e$ $=$ electronic charge)
1 $\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
2 $\frac{-\mathrm{eh}}{\pi \mathrm{m}}$
3 $\frac{\mathrm{eh}}{2 \pi \mathrm{m}}$
4 $\frac{2 \mathrm{eh}}{\pi \mathrm{m}}$
Explanation:
A According to Bohr's theory- $\operatorname{mvR}=\frac{\mathrm{h}}{2 \pi}$ or $\quad \mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}$ We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}$ And, $\quad I=\frac{|\mathrm{e}| \overrightarrow{\mathrm{A}}}{\mathrm{T}}$ Magnetic dipole moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M}=\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}} \times \pi \mathrm{R}^{2}$ $\mathrm{M}=\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
MHT-CET 2020
Moving Charges & Magnetism
153907
An electron revolving in circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment. ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{4}$
2 $\frac{M}{2}$
3 $\mathrm{M}$
4 $2 \mathrm{M}$
Explanation:
D We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And $\quad \mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}=\mathrm{ef}$ $\left(\because \mathrm{f}=\frac{1}{\mathrm{~T}}\right)$ $\mathrm{I} \propto \mathrm{f}$ Therefore, $\mathrm{M} \propto \mathrm{I} \propto \mathrm{f}$ Then, $\frac{M_{1}}{M_{2}}=\frac{f_{1}}{f_{2}}$ $\left(\because \mathrm{M}_{1}=\mathrm{M}, \mathrm{f}_{2}=2 \mathrm{f}_{1}\right)$ $\mathrm{M}_{2}=2 \mathrm{M}$
MHT-CET 2020
Moving Charges & Magnetism
153909
A circular and a square coil is prepared from two identical metal wires and a current is passed through them. Ratio of magnetic dipole moment associated with circular coil to that with square coil is
1 $\frac{2}{\pi}$
2 $\frac{4}{\pi}$
3 $\frac{\pi}{2}$
4 $\pi$
Explanation:
B Let the length of the wire $\mathrm{L}$ 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 - $\frac{\left(\mathrm{M}_{2}\right)_{\text {circular loop }}}{\left(\mathrm{M}_{1}\right)_{\text {square }}}=\frac{\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}}{\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}}=\frac{4}{\pi}$
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Moving Charges & Magnetism
153901
A particle of charge ' $q$ ' and mass ' $m$ ' moves in a circular orbit of radius ' $r$ ' with angular speed ' $\omega$ '. The ratio of the magnitude of its magnetic moment to that its angular momentum is
1 $\frac{q \omega r^{2}}{2 m}$
2 $\frac{q \omega r^{2}}{2}$
3 $\frac{q \omega}{2 m r^{2}}$
4 $\frac{\mathrm{q}}{2 \mathrm{~m}}$
Explanation:
D Given, radius of orbit $=r$, magnetic moment $=\mathrm{M}$, angular speed $=\theta$ $\therefore \quad \mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}} \quad$ (circumference of circle $=2 \pi \mathrm{r}$ ) And, $\quad I=\frac{q}{t}=\frac{q v}{2 \pi r}$ $\therefore$ Magnetic moment $(\mathrm{M})=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{qvr}}{2}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ Dividing equation (i) and (ii), we get- $\frac{M}{L}=\frac{\frac{q v r}{2}}{m v r}$ $\frac{M}{L}=\frac{q}{2 m}$
MHT-CET 2020
Moving Charges & Magnetism
153902
Which one of the following proportionality represents the relation between orbital magnetic moment $M_{0}$ and orbital angular momentum $L_{0}$ of an electron?
D We know that, magnetic moment (M) = IA And, $t=\frac{2 \pi r}{v}$ $\therefore \quad \mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{q}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ So, $\quad \mathrm{M}_{0}=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2} \quad\left(\pi \mathrm{r}^{2}=\right.$ area of circle $)$ $\mathrm{M}_{0}=\frac{\mathrm{evr}}{2} \quad \ldots . \text { (i) } \quad(\because \mathrm{q}=\mathrm{e})$ Angular momentum $\left(\mathrm{L}_{0}\right)=\mathrm{mvr}$ From equation (ii), we get- $\mathrm{vr}=\frac{\mathrm{L}_{0}}{\mathrm{~m}}$ Putting the value of 'vr' in equation (i), we get- $\mathrm{M}_{0}=\frac{\mathrm{e}}{2} \times \frac{\mathrm{L}_{0}}{\mathrm{~m}}$ $\therefore \quad \mathrm{M}_{0} \propto \mathrm{L}_{0}$
MHT-CET 2020
Moving Charges & Magnetism
153906
An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is $(m=$ mass of electron, $h=$ Planck's constant, $e$ $=$ electronic charge)
1 $\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
2 $\frac{-\mathrm{eh}}{\pi \mathrm{m}}$
3 $\frac{\mathrm{eh}}{2 \pi \mathrm{m}}$
4 $\frac{2 \mathrm{eh}}{\pi \mathrm{m}}$
Explanation:
A According to Bohr's theory- $\operatorname{mvR}=\frac{\mathrm{h}}{2 \pi}$ or $\quad \mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}$ We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}$ And, $\quad I=\frac{|\mathrm{e}| \overrightarrow{\mathrm{A}}}{\mathrm{T}}$ Magnetic dipole moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M}=\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}} \times \pi \mathrm{R}^{2}$ $\mathrm{M}=\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
MHT-CET 2020
Moving Charges & Magnetism
153907
An electron revolving in circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment. ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{4}$
2 $\frac{M}{2}$
3 $\mathrm{M}$
4 $2 \mathrm{M}$
Explanation:
D We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And $\quad \mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}=\mathrm{ef}$ $\left(\because \mathrm{f}=\frac{1}{\mathrm{~T}}\right)$ $\mathrm{I} \propto \mathrm{f}$ Therefore, $\mathrm{M} \propto \mathrm{I} \propto \mathrm{f}$ Then, $\frac{M_{1}}{M_{2}}=\frac{f_{1}}{f_{2}}$ $\left(\because \mathrm{M}_{1}=\mathrm{M}, \mathrm{f}_{2}=2 \mathrm{f}_{1}\right)$ $\mathrm{M}_{2}=2 \mathrm{M}$
MHT-CET 2020
Moving Charges & Magnetism
153909
A circular and a square coil is prepared from two identical metal wires and a current is passed through them. Ratio of magnetic dipole moment associated with circular coil to that with square coil is
1 $\frac{2}{\pi}$
2 $\frac{4}{\pi}$
3 $\frac{\pi}{2}$
4 $\pi$
Explanation:
B Let the length of the wire $\mathrm{L}$ 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 - $\frac{\left(\mathrm{M}_{2}\right)_{\text {circular loop }}}{\left(\mathrm{M}_{1}\right)_{\text {square }}}=\frac{\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}}{\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}}=\frac{4}{\pi}$
153901
A particle of charge ' $q$ ' and mass ' $m$ ' moves in a circular orbit of radius ' $r$ ' with angular speed ' $\omega$ '. The ratio of the magnitude of its magnetic moment to that its angular momentum is
1 $\frac{q \omega r^{2}}{2 m}$
2 $\frac{q \omega r^{2}}{2}$
3 $\frac{q \omega}{2 m r^{2}}$
4 $\frac{\mathrm{q}}{2 \mathrm{~m}}$
Explanation:
D Given, radius of orbit $=r$, magnetic moment $=\mathrm{M}$, angular speed $=\theta$ $\therefore \quad \mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}} \quad$ (circumference of circle $=2 \pi \mathrm{r}$ ) And, $\quad I=\frac{q}{t}=\frac{q v}{2 \pi r}$ $\therefore$ Magnetic moment $(\mathrm{M})=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{qvr}}{2}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ Dividing equation (i) and (ii), we get- $\frac{M}{L}=\frac{\frac{q v r}{2}}{m v r}$ $\frac{M}{L}=\frac{q}{2 m}$
MHT-CET 2020
Moving Charges & Magnetism
153902
Which one of the following proportionality represents the relation between orbital magnetic moment $M_{0}$ and orbital angular momentum $L_{0}$ of an electron?
D We know that, magnetic moment (M) = IA And, $t=\frac{2 \pi r}{v}$ $\therefore \quad \mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{q}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ So, $\quad \mathrm{M}_{0}=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2} \quad\left(\pi \mathrm{r}^{2}=\right.$ area of circle $)$ $\mathrm{M}_{0}=\frac{\mathrm{evr}}{2} \quad \ldots . \text { (i) } \quad(\because \mathrm{q}=\mathrm{e})$ Angular momentum $\left(\mathrm{L}_{0}\right)=\mathrm{mvr}$ From equation (ii), we get- $\mathrm{vr}=\frac{\mathrm{L}_{0}}{\mathrm{~m}}$ Putting the value of 'vr' in equation (i), we get- $\mathrm{M}_{0}=\frac{\mathrm{e}}{2} \times \frac{\mathrm{L}_{0}}{\mathrm{~m}}$ $\therefore \quad \mathrm{M}_{0} \propto \mathrm{L}_{0}$
MHT-CET 2020
Moving Charges & Magnetism
153906
An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is $(m=$ mass of electron, $h=$ Planck's constant, $e$ $=$ electronic charge)
1 $\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
2 $\frac{-\mathrm{eh}}{\pi \mathrm{m}}$
3 $\frac{\mathrm{eh}}{2 \pi \mathrm{m}}$
4 $\frac{2 \mathrm{eh}}{\pi \mathrm{m}}$
Explanation:
A According to Bohr's theory- $\operatorname{mvR}=\frac{\mathrm{h}}{2 \pi}$ or $\quad \mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}$ We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}$ And, $\quad I=\frac{|\mathrm{e}| \overrightarrow{\mathrm{A}}}{\mathrm{T}}$ Magnetic dipole moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M}=\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}} \times \pi \mathrm{R}^{2}$ $\mathrm{M}=\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
MHT-CET 2020
Moving Charges & Magnetism
153907
An electron revolving in circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment. ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{4}$
2 $\frac{M}{2}$
3 $\mathrm{M}$
4 $2 \mathrm{M}$
Explanation:
D We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And $\quad \mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}=\mathrm{ef}$ $\left(\because \mathrm{f}=\frac{1}{\mathrm{~T}}\right)$ $\mathrm{I} \propto \mathrm{f}$ Therefore, $\mathrm{M} \propto \mathrm{I} \propto \mathrm{f}$ Then, $\frac{M_{1}}{M_{2}}=\frac{f_{1}}{f_{2}}$ $\left(\because \mathrm{M}_{1}=\mathrm{M}, \mathrm{f}_{2}=2 \mathrm{f}_{1}\right)$ $\mathrm{M}_{2}=2 \mathrm{M}$
MHT-CET 2020
Moving Charges & Magnetism
153909
A circular and a square coil is prepared from two identical metal wires and a current is passed through them. Ratio of magnetic dipole moment associated with circular coil to that with square coil is
1 $\frac{2}{\pi}$
2 $\frac{4}{\pi}$
3 $\frac{\pi}{2}$
4 $\pi$
Explanation:
B Let the length of the wire $\mathrm{L}$ 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 - $\frac{\left(\mathrm{M}_{2}\right)_{\text {circular loop }}}{\left(\mathrm{M}_{1}\right)_{\text {square }}}=\frac{\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}}{\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}}=\frac{4}{\pi}$
153901
A particle of charge ' $q$ ' and mass ' $m$ ' moves in a circular orbit of radius ' $r$ ' with angular speed ' $\omega$ '. The ratio of the magnitude of its magnetic moment to that its angular momentum is
1 $\frac{q \omega r^{2}}{2 m}$
2 $\frac{q \omega r^{2}}{2}$
3 $\frac{q \omega}{2 m r^{2}}$
4 $\frac{\mathrm{q}}{2 \mathrm{~m}}$
Explanation:
D Given, radius of orbit $=r$, magnetic moment $=\mathrm{M}$, angular speed $=\theta$ $\therefore \quad \mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}} \quad$ (circumference of circle $=2 \pi \mathrm{r}$ ) And, $\quad I=\frac{q}{t}=\frac{q v}{2 \pi r}$ $\therefore$ Magnetic moment $(\mathrm{M})=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{qvr}}{2}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ Dividing equation (i) and (ii), we get- $\frac{M}{L}=\frac{\frac{q v r}{2}}{m v r}$ $\frac{M}{L}=\frac{q}{2 m}$
MHT-CET 2020
Moving Charges & Magnetism
153902
Which one of the following proportionality represents the relation between orbital magnetic moment $M_{0}$ and orbital angular momentum $L_{0}$ of an electron?
D We know that, magnetic moment (M) = IA And, $t=\frac{2 \pi r}{v}$ $\therefore \quad \mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{q}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ So, $\quad \mathrm{M}_{0}=\mathrm{IA}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2} \quad\left(\pi \mathrm{r}^{2}=\right.$ area of circle $)$ $\mathrm{M}_{0}=\frac{\mathrm{evr}}{2} \quad \ldots . \text { (i) } \quad(\because \mathrm{q}=\mathrm{e})$ Angular momentum $\left(\mathrm{L}_{0}\right)=\mathrm{mvr}$ From equation (ii), we get- $\mathrm{vr}=\frac{\mathrm{L}_{0}}{\mathrm{~m}}$ Putting the value of 'vr' in equation (i), we get- $\mathrm{M}_{0}=\frac{\mathrm{e}}{2} \times \frac{\mathrm{L}_{0}}{\mathrm{~m}}$ $\therefore \quad \mathrm{M}_{0} \propto \mathrm{L}_{0}$
MHT-CET 2020
Moving Charges & Magnetism
153906
An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is $(m=$ mass of electron, $h=$ Planck's constant, $e$ $=$ electronic charge)
1 $\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
2 $\frac{-\mathrm{eh}}{\pi \mathrm{m}}$
3 $\frac{\mathrm{eh}}{2 \pi \mathrm{m}}$
4 $\frac{2 \mathrm{eh}}{\pi \mathrm{m}}$
Explanation:
A According to Bohr's theory- $\operatorname{mvR}=\frac{\mathrm{h}}{2 \pi}$ or $\quad \mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}$ We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}$ And, $\quad I=\frac{|\mathrm{e}| \overrightarrow{\mathrm{A}}}{\mathrm{T}}$ Magnetic dipole moment $(\mathrm{M})=\mathrm{I} \times \mathrm{A}$ $\mathrm{M}=\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}} \times \pi \mathrm{R}^{2}$ $\mathrm{M}=\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
MHT-CET 2020
Moving Charges & Magnetism
153907
An electron revolving in circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment. ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{4}$
2 $\frac{M}{2}$
3 $\mathrm{M}$
4 $2 \mathrm{M}$
Explanation:
D We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And $\quad \mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}=\mathrm{ef}$ $\left(\because \mathrm{f}=\frac{1}{\mathrm{~T}}\right)$ $\mathrm{I} \propto \mathrm{f}$ Therefore, $\mathrm{M} \propto \mathrm{I} \propto \mathrm{f}$ Then, $\frac{M_{1}}{M_{2}}=\frac{f_{1}}{f_{2}}$ $\left(\because \mathrm{M}_{1}=\mathrm{M}, \mathrm{f}_{2}=2 \mathrm{f}_{1}\right)$ $\mathrm{M}_{2}=2 \mathrm{M}$
MHT-CET 2020
Moving Charges & Magnetism
153909
A circular and a square coil is prepared from two identical metal wires and a current is passed through them. Ratio of magnetic dipole moment associated with circular coil to that with square coil is
1 $\frac{2}{\pi}$
2 $\frac{4}{\pi}$
3 $\frac{\pi}{2}$
4 $\pi$
Explanation:
B Let the length of the wire $\mathrm{L}$ 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 - $\frac{\left(\mathrm{M}_{2}\right)_{\text {circular loop }}}{\left(\mathrm{M}_{1}\right)_{\text {square }}}=\frac{\mathrm{I} \times 2 \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}}{\mathrm{I} \times\left(\frac{\mathrm{L}}{4}\right)^{2}}=\frac{4}{\pi}$