358422
A solenoid of inductance ' \(L\) ' carrying a certain current is linked with a total magnetic flux \(\phi\). Now it is connected to a condenser with which it shares half of its initial energy. The total flux now linked with the solenoid is
1 \(\phi / 2\)
2 \(\phi / \sqrt{2}\)
3 \(\phi / 2 \sqrt{2}\)
4 \(\phi / 4\)
Explanation:
Flux through a solenoid is \(\phi=L i\) Energy stored in a solenoid is \(U=\dfrac{1}{2} L i^{2}=\dfrac{\phi^{2}}{2 L}\) \(\dfrac{U_{1}}{U_{2}}=\left(\dfrac{\phi_{1}}{\phi_{2}}\right)^{2} \Rightarrow U_{2}=U_{1} / 2 \Rightarrow \phi_{2}=\dfrac{\phi_{1}}{\sqrt{2}}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358423
A long solenoid with 40 turns per \(cm\) carries a current of \(1\;A\). The magnetic energy stored per unit volume is \(J/{m^3}.\)
358424
A solenoid of inductance \(2\,H\) carries a current of \(1\;\,A\). What is the magnetic energy stored in the solenoid?
1 \(4\;\,J\)
2 \(2\;\,J\)
3 \(5\;\,J\)
4 \(1\;\,J\)
Explanation:
Here, Inductance,\(L = 2H\), current \(I = IA\) Magnetic energy stored in the solenoid is \(U = \frac{1}{2}L{I^2} = \frac{1}{2} \times 2 \times {(1)^2} = 1\;J\)
KCET - 2014
PHXII06:ELECTROMAGNETIC INDUCTION
358425
The magnetic energy stored in an inductor of inductance \(4\,\mu H\) carrying a current of \(2\;A\) is:
1 \(4\;mJ\)
2 \(8\;mJ\)
3 \(8\,\mu J\)
4 \(4\,\mu J\)
Explanation:
Energy \(U=\dfrac{1}{2} L i^{2}\) \(U=\dfrac{1}{2} 4 \times 10^{-6} \times 2^{2}=8 \times 10^{-6} J=8 \mu J\) Correct option is (3).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII06:ELECTROMAGNETIC INDUCTION
358422
A solenoid of inductance ' \(L\) ' carrying a certain current is linked with a total magnetic flux \(\phi\). Now it is connected to a condenser with which it shares half of its initial energy. The total flux now linked with the solenoid is
1 \(\phi / 2\)
2 \(\phi / \sqrt{2}\)
3 \(\phi / 2 \sqrt{2}\)
4 \(\phi / 4\)
Explanation:
Flux through a solenoid is \(\phi=L i\) Energy stored in a solenoid is \(U=\dfrac{1}{2} L i^{2}=\dfrac{\phi^{2}}{2 L}\) \(\dfrac{U_{1}}{U_{2}}=\left(\dfrac{\phi_{1}}{\phi_{2}}\right)^{2} \Rightarrow U_{2}=U_{1} / 2 \Rightarrow \phi_{2}=\dfrac{\phi_{1}}{\sqrt{2}}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358423
A long solenoid with 40 turns per \(cm\) carries a current of \(1\;A\). The magnetic energy stored per unit volume is \(J/{m^3}.\)
358424
A solenoid of inductance \(2\,H\) carries a current of \(1\;\,A\). What is the magnetic energy stored in the solenoid?
1 \(4\;\,J\)
2 \(2\;\,J\)
3 \(5\;\,J\)
4 \(1\;\,J\)
Explanation:
Here, Inductance,\(L = 2H\), current \(I = IA\) Magnetic energy stored in the solenoid is \(U = \frac{1}{2}L{I^2} = \frac{1}{2} \times 2 \times {(1)^2} = 1\;J\)
KCET - 2014
PHXII06:ELECTROMAGNETIC INDUCTION
358425
The magnetic energy stored in an inductor of inductance \(4\,\mu H\) carrying a current of \(2\;A\) is:
1 \(4\;mJ\)
2 \(8\;mJ\)
3 \(8\,\mu J\)
4 \(4\,\mu J\)
Explanation:
Energy \(U=\dfrac{1}{2} L i^{2}\) \(U=\dfrac{1}{2} 4 \times 10^{-6} \times 2^{2}=8 \times 10^{-6} J=8 \mu J\) Correct option is (3).
358422
A solenoid of inductance ' \(L\) ' carrying a certain current is linked with a total magnetic flux \(\phi\). Now it is connected to a condenser with which it shares half of its initial energy. The total flux now linked with the solenoid is
1 \(\phi / 2\)
2 \(\phi / \sqrt{2}\)
3 \(\phi / 2 \sqrt{2}\)
4 \(\phi / 4\)
Explanation:
Flux through a solenoid is \(\phi=L i\) Energy stored in a solenoid is \(U=\dfrac{1}{2} L i^{2}=\dfrac{\phi^{2}}{2 L}\) \(\dfrac{U_{1}}{U_{2}}=\left(\dfrac{\phi_{1}}{\phi_{2}}\right)^{2} \Rightarrow U_{2}=U_{1} / 2 \Rightarrow \phi_{2}=\dfrac{\phi_{1}}{\sqrt{2}}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358423
A long solenoid with 40 turns per \(cm\) carries a current of \(1\;A\). The magnetic energy stored per unit volume is \(J/{m^3}.\)
358424
A solenoid of inductance \(2\,H\) carries a current of \(1\;\,A\). What is the magnetic energy stored in the solenoid?
1 \(4\;\,J\)
2 \(2\;\,J\)
3 \(5\;\,J\)
4 \(1\;\,J\)
Explanation:
Here, Inductance,\(L = 2H\), current \(I = IA\) Magnetic energy stored in the solenoid is \(U = \frac{1}{2}L{I^2} = \frac{1}{2} \times 2 \times {(1)^2} = 1\;J\)
KCET - 2014
PHXII06:ELECTROMAGNETIC INDUCTION
358425
The magnetic energy stored in an inductor of inductance \(4\,\mu H\) carrying a current of \(2\;A\) is:
1 \(4\;mJ\)
2 \(8\;mJ\)
3 \(8\,\mu J\)
4 \(4\,\mu J\)
Explanation:
Energy \(U=\dfrac{1}{2} L i^{2}\) \(U=\dfrac{1}{2} 4 \times 10^{-6} \times 2^{2}=8 \times 10^{-6} J=8 \mu J\) Correct option is (3).
358422
A solenoid of inductance ' \(L\) ' carrying a certain current is linked with a total magnetic flux \(\phi\). Now it is connected to a condenser with which it shares half of its initial energy. The total flux now linked with the solenoid is
1 \(\phi / 2\)
2 \(\phi / \sqrt{2}\)
3 \(\phi / 2 \sqrt{2}\)
4 \(\phi / 4\)
Explanation:
Flux through a solenoid is \(\phi=L i\) Energy stored in a solenoid is \(U=\dfrac{1}{2} L i^{2}=\dfrac{\phi^{2}}{2 L}\) \(\dfrac{U_{1}}{U_{2}}=\left(\dfrac{\phi_{1}}{\phi_{2}}\right)^{2} \Rightarrow U_{2}=U_{1} / 2 \Rightarrow \phi_{2}=\dfrac{\phi_{1}}{\sqrt{2}}\)
PHXII06:ELECTROMAGNETIC INDUCTION
358423
A long solenoid with 40 turns per \(cm\) carries a current of \(1\;A\). The magnetic energy stored per unit volume is \(J/{m^3}.\)
358424
A solenoid of inductance \(2\,H\) carries a current of \(1\;\,A\). What is the magnetic energy stored in the solenoid?
1 \(4\;\,J\)
2 \(2\;\,J\)
3 \(5\;\,J\)
4 \(1\;\,J\)
Explanation:
Here, Inductance,\(L = 2H\), current \(I = IA\) Magnetic energy stored in the solenoid is \(U = \frac{1}{2}L{I^2} = \frac{1}{2} \times 2 \times {(1)^2} = 1\;J\)
KCET - 2014
PHXII06:ELECTROMAGNETIC INDUCTION
358425
The magnetic energy stored in an inductor of inductance \(4\,\mu H\) carrying a current of \(2\;A\) is:
1 \(4\;mJ\)
2 \(8\;mJ\)
3 \(8\,\mu J\)
4 \(4\,\mu J\)
Explanation:
Energy \(U=\dfrac{1}{2} L i^{2}\) \(U=\dfrac{1}{2} 4 \times 10^{-6} \times 2^{2}=8 \times 10^{-6} J=8 \mu J\) Correct option is (3).