NEET-XII-Physics
38: Electromagnetic Induction
- #22-athe average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part
(a). (b) the average emf in a full turn and (c) the net charge displaced in part
(a).Ans : When the coil is perpendicular to the field:
Ï•1 = nBA
When the coil goes through the half turn:
ϕ2 = nBA cos 180° = -nBA
∴ Δϕ = 2nBA
When the coil undergoes 300 revolutions in 1 minute, the angle swept by the coil is
300 × 2π rad/min = 10π rad/s
10Ï€ rad is swept in 1 s.
Ï€ rad is swept in `` \left(\frac{1}{10\,\mathrm{\,\pi \,}}\right)\,\mathrm{\,\pi \,}=\frac{1}{10}\,\mathrm{\,s\,}``
`` e=\frac{d\,\mathrm{\,\varphi \,}}{dt}=\frac{2nBA}{dt}``
`` =\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{1/10}``
`` =2\times {10}^{-3}\,\mathrm{\,V\,}`` (b) ϕ1 = nBA, ϕ2 = nBA (θ = 360°)
Δϕ = 0, thus emf induced will be zero. (c) The current flowing in the coil is given by
`` i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}``
= 0.5 × 10-3 = 5 × 10-4 A
Hence, the net charge is given by
Q = idt = 5 × 10-4 × `` \frac{1}{10}``
= 5 × 10-5 C
Page No 307: (b) ϕ1 = nBA, ϕ2 = nBA (θ = 360°)
Δϕ = 0, thus emf induced will be zero. (c) The current flowing in the coil is given by
`` i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}``
= 0.5 × 10-3 = 5 × 10-4 A
Hence, the net charge is given by
Q = idt = 5 × 10-4 × `` \frac{1}{10}``
= 5 × 10-5 C
Page No 307:
- #22-bthe average emf in a full turn andAns : ϕ1 = nBA, ϕ2 = nBA (θ = 360°)
Δϕ = 0, thus emf induced will be zero.
- #22-cthe net charge displaced in part
(a).Ans : The current flowing in the coil is given by
`` i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}``
= 0.5 × 10-3 = 5 × 10-4 A
Hence, the net charge is given by
Q = idt = 5 × 10-4 × `` \frac{1}{10}``
= 5 × 10-5 C
Page No 307: