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The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance:
- a)0.138 H
- b)138.88 H
- c)1.389 H
- d)13.89 H
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The magnetic potential energy stored in a certain inductor is 25 mJ, w...
Concept:
- The coil which stores magnetic energy in a magnetic field is called an inductor.
- The property of an inductor which causes the emf to generate by a change in electric current is called as inductance of the inductor.
- The SI unit of inductance is Henry (H).
- The magnetic potential energy stored in an inductor is given by,
Where L is inductance of the inductor and I is current flowing.
Explanation:
Given that:
Magnetic potential energy (U) = 25 mJ = 25 x 10-3 J
Current (I) = 60 mA = 60 x 10-3A
Use the formula:
Given that:
Magnetic potential energy (U) = 25 mJ = 25 x 10-3 J
Current (I) = 60 mA = 60 x 10-3A
Use the formula:
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The magnetic potential energy stored in a certain inductor is 25 mJ, w...
Calculating Inductance:
- The magnetic potential energy stored in an inductor is given by the formula:
\[W = \frac{1}{2}LI^2\]
where:
W = magnetic potential energy stored (25 mJ = 25 x 10^-3 J)
L = inductance of the inductor (to be calculated)
I = current in the inductor (60 mA = 60 x 10^-3 A)
- Substituting the given values into the formula:
\[25 x 10^-3 J = \frac{1}{2}L(60 x 10^-3 A)^2\]
Solving for Inductance:
- Rearranging the formula to solve for inductance:
\[L = \frac{2W}{I^2}\]
\[L = \frac{2(25 x 10^-3)}{(60 x 10^-3)^2}\]
\[L = \frac{50 x 10^-3}{3600 x 10^-6}\]
\[L = \frac{50}{3600} H\]
\[L = 13.89 H\]
Therefore, the inductance of the inductor is 13.89 H, which corresponds to option 'D'.
- The magnetic potential energy stored in an inductor is given by the formula:
\[W = \frac{1}{2}LI^2\]
where:
W = magnetic potential energy stored (25 mJ = 25 x 10^-3 J)
L = inductance of the inductor (to be calculated)
I = current in the inductor (60 mA = 60 x 10^-3 A)
- Substituting the given values into the formula:
\[25 x 10^-3 J = \frac{1}{2}L(60 x 10^-3 A)^2\]
Solving for Inductance:
- Rearranging the formula to solve for inductance:
\[L = \frac{2W}{I^2}\]
\[L = \frac{2(25 x 10^-3)}{(60 x 10^-3)^2}\]
\[L = \frac{50 x 10^-3}{3600 x 10^-6}\]
\[L = \frac{50}{3600} H\]
\[L = 13.89 H\]
Therefore, the inductance of the inductor is 13.89 H, which corresponds to option 'D'.
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The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance:a)0.138 Hb)138.88 Hc)1.389 Hd)13.89 HCorrect answer is option 'D'. Can you explain this answer?
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