Electromagnetic Fields Test 2

Result

Electromagnetic Fields Test 2

Score
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Rank
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Weekly Quiz Competition

Question 1
1 / -0

The energy stored in the magnetic field of the solenoid of length 40 cm and 8 cm diameter wound with 1000 turns of wire carrying a current of 10 A is ________ J (μ_r = 1)

A
0.75-0.85

B
0.754-0.851

C
0.757-0.856

D
0.759-0.852

Solution

Length of Solenoid = 40 cm
Diameter of solenoid = 8 cm
No. of turns = 1000
Current carrying = 10 A
The energy stored in the magnetic field \( = \frac{1}{2}L\;{i^2}\)
The the inductance of the solenoid \(L = \frac{{\mu {N^2}a}}{\ell }\)
\(L = \frac{{4\pi \times {{10}^{ - 7}} \times 1 \times {{1000}^2} \times \pi \times {{\left( {4 \times {{10}^{ - 2}}} \right)}^2}}}{{40 \times {{10}^{ - 2}}}}\)
⇒ L = 15.791 mH
Energy stored \( = \frac{1}{2}L\;{I^2} = \frac{1}{2} \times 15.791 \times {10^{ - 3}} \times {10^2}\)
= 0.789 J
Question 2
1 / -0

The toroidal core has mean radius 10 cm and circular cross-section with a = 1 cm if the core is made of steel (μ = 1000 μ₀) and has a coil will 200 turns, then the amount of current that will produce a flux of 0.5 mWb in the core is ________ A.

A
3.9-4

B
3.94-41

C
3.97-46

D
3.99-42

Solution

Radius of core = 10 cm
Cross sectional area of core a = 1 cm
μ_r = 1000
Number of turns = 200
Flux = 0.5 × 10^-3 wb
We know flux \(\phi = \frac{{mmf}}{s} = \frac{{NI}}{s}\)
Where \(s = \frac{l}{{{\mu _0}{\mu _r}A}}\;AT/wb\)
\(I = \frac{{\phi s}}{N} = \frac{{\phi l}}{{{u_0}{u_r}AN}}\) Where (l = 2πr)
\(I = \frac{{0.5 \times {{10}^{ - 3}} \times \left( {2\pi r} \right) \times 10 \times {{10}^{ - 2}}}}{{4\pi \times {{10}^{ - 7}} \times {{10}^3} \times 200 \times \pi \times {{10}^{ - 4}}}}\;A\)
I = 3.978 A