Electromagnetic theory miscellaneous
- The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter wound with 1000 turns of wire carrying a current of 10 A is—
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Since we know that,
L = N2 × ρ
given
N = 1000 turns diameter
= 3 cm, L = 30 cm.ρ = µ0A = 4π × 10–7 × π × (1.5 × 10–2)2 L 0.3
= 29.6088 × 10–10
L = 29.6088 × 10–10 × (103)2
= 29.6088 × 10–4 H
where,
L = length of solenoid
W = 1/2 LI2
= 1/2 × 29.6088 × 10–4 × (10)2
= 0.148 ≈ 0.15 J.Correct Option: B
Since we know that,
L = N2 × ρ
given
N = 1000 turns diameter
= 3 cm, L = 30 cm.ρ = µ0A = 4π × 10–7 × π × (1.5 × 10–2)2 L 0.3
= 29.6088 × 10–10
L = 29.6088 × 10–10 × (103)2
= 29.6088 × 10–4 H
where,
L = length of solenoid
W = 1/2 LI2
= 1/2 × 29.6088 × 10–4 × (10)2
= 0.148 ≈ 0.15 J.
- If V, w, q stand for voltage, energy and charge, then V can be expressed as—
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Since voltage is, broadly saying work per unit charge, we can express V as w/q, or V as dw/dq.
Correct Option: B
Since voltage is, broadly saying work per unit charge, we can express V as w/q, or V as dw/dq.
- Inside a hollow conducting sphere—
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For any point inside the conducting sphere, the charge enclosed by the concentric sphere is zero and hence E is zero.
Correct Option: A
For any point inside the conducting sphere, the charge enclosed by the concentric sphere is zero and hence E is zero.
- The expression for B, given that in free space E = 15 cos (6π × 108 t – 2πz) ix V/m is—
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The given wave is travelling in the + z direction (because of the presence of the term – 2 πz) and E is entirely in the x-direction. Now E and H direction gives the direction of travel of tower. It is concluded, therefore, that H must be in + y direction.
ix × iy = iz
Further, we know that| H | = |E| and B = µ0 H η η = √ µ0 or | B | = µ0 |E| ε0 √µ0/ε0
or
B = E/V= ∵ = V 1 √µ0∈0
orB = 15 cos (6π × 108 t – 2πz) iy 3 × 108
or
B = 5 × 10-8 cos (6π × 108 t – 2π z) iy Wb/m2Correct Option: A
The given wave is travelling in the + z direction (because of the presence of the term – 2 πz) and E is entirely in the x-direction. Now E and H direction gives the direction of travel of tower. It is concluded, therefore, that H must be in + y direction.
ix × iy = iz
Further, we know that| H | = |E| and B = µ0 H η η = √ µ0 or | B | = µ0 |E| ε0 √µ0/ε0
or
B = E/V= ∵ = V 1 √µ0∈0
orB = 15 cos (6π × 108 t – 2πz) iy 3 × 108
or
B = 5 × 10-8 cos (6π × 108 t – 2π z) iy Wb/m2
- Following equations hold for time-varying fields—
(i)∇ × E = − ∂B ∂t
(ii)E = − ∇ − ∂A ∂t
(iii)∇2V + ∂ (∇⋅A) = −ρv ∂t ε
(iv)→ → B = ∇ × A
(v)→ → ∇ × H = J + ∂E ∂t
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Both V and →A are completely defined and thus can be evaluated.
Correct Option: A
Both V and →A are completely defined and thus can be evaluated.
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