Question
Question: A solenoid of self-inductance \[1.2{\text{ }}H\] is in series with a tangent galvanometer of reducti...
A solenoid of self-inductance 1.2 H is in series with a tangent galvanometer of reduction factor 0.9A. They are connected to a battery and the tangent galvanometer shows a deflection of 53∘. The energy stored in the magnetic field of the solenoid is:
(tan 53∘ = 34)
A. 0.864J
B. 0.72J
C. 0.173J
D. 1.44J
Solution
To solve the question, i.e., to find the energy stored in the magnetic field of the solenoid, we will start with finding the current, because to find the energy we will need the value of current, so after finding the value of current using reduction factor and deflection, we will use the energy formula and get our required answer.
Complete step by step answer: We have been given a solenoid of self-inductance 1.2 H is in series with a tangent galvanometer of reduction factor 0.9A. They are connected to a battery and the tangent galvanometer shows a deflection of 53∘. We need to find the energy stored in the magnetic field of the solenoid is:
(tan 53∘ = 34)
We know that, current = reduction factor × deflection
Here, reduction factor = 0.9, deflection = tan 53∘
On substituting the value in the above formula, we get