Qstnid -Iv-2-the Flux Of Magnetic Field Through A Closed Conducting Loop Changes With Time According To The Equation, Φ = At<sup>2</sup> + Bt + C. (A) Write The Si Units Of A, B And C. (B) If The Magnitudes Of A, B And C Are 0.20, 0.40 And 0.60 Respectively, Find The Induced Emf At T = 2 S. (A) Write The Si Units Of A, B And C. (B) If The Magnitudes Of A, B And C Are 0.20, 0.40 And 0.60 Respectively, Find The Induced Emf At T = 2 S. (B) If The Magnitudes Of A, B And C Are 0.20, 0.40 And 0.60 Respectively, Find The Induced Emf At T = 2 S. - 38: Electromagnetic Induction

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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 4

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#2
The flux of magnetic field through a closed conducting loop changes with time according to the equation, Φ = at² + bt + c. (a) Write the SI units of a, b and c. (b) If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s. (a) Write the SI units of a, b and c. (b) If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s. (b) If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s.
Ans : According to the principle of homogeneity of dimensions, the dimensions of each term on both the sides of a correct equation must be the same.
Now,
ϕ = at² + bt + c (a) The dimensions of the quantities at², bt, c and ϕ must be the same.
Thus, the units of the quantities are as follows:
`` a=\left(\frac{\,\mathrm{\,\varphi \,}}{{t}^{2}}\right)=\left[\frac{\,\mathrm{\,\varphi \,}/t}{t}\right]=\frac{\,\mathrm{\,Volt\,}}{\,\mathrm{\,s\,}}``
`` b=\left[\frac{\,\mathrm{\,\varphi \,}}{t}\right]=\,\mathrm{\,Volt\,}``
`` c=\left[\,\mathrm{\,\varphi \,}\right]=\,\mathrm{\,Weber\,}`` (b) The emf is written as:
`` E=\frac{d\varphi }{dt}`` = 2at + b = 2 × 0.2 × 2 + 0.4 (∵ a = 0.2, b = 0.4 and c = 0.6)
On substituting t = 2 s, we get
E = 0.8 + 0.4
= 1.2 V
Page No 306: (a) The dimensions of the quantities at², bt, c and ϕ must be the same.
Thus, the units of the quantities are as follows:
`` a=\left(\frac{\,\mathrm{\,\varphi \,}}{{t}^{2}}\right)=\left[\frac{\,\mathrm{\,\varphi \,}/t}{t}\right]=\frac{\,\mathrm{\,Volt\,}}{\,\mathrm{\,s\,}}``
`` b=\left[\frac{\,\mathrm{\,\varphi \,}}{t}\right]=\,\mathrm{\,Volt\,}``
`` c=\left[\,\mathrm{\,\varphi \,}\right]=\,\mathrm{\,Weber\,}`` (b) The emf is written as:
`` E=\frac{d\varphi }{dt}`` = 2at + b = 2 × 0.2 × 2 + 0.4 (∵ a = 0.2, b = 0.4 and c = 0.6)
On substituting t = 2 s, we get
E = 0.8 + 0.4
= 1.2 V
Page No 306: (b) The emf is written as:
`` E=\frac{d\varphi }{dt}`` = 2at + b = 2 × 0.2 × 2 + 0.4 (∵ a = 0.2, b = 0.4 and c = 0.6)
On substituting t = 2 s, we get
E = 0.8 + 0.4
= 1.2 V
Page No 306:

#2-a
Write the SI units of a, b and c. (b) If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s.
Ans : The dimensions of the quantities at², bt, c and ϕ must be the same.
Thus, the units of the quantities are as follows:
`` a=\left(\frac{\,\mathrm{\,\varphi \,}}{{t}^{2}}\right)=\left[\frac{\,\mathrm{\,\varphi \,}/t}{t}\right]=\frac{\,\mathrm{\,Volt\,}}{\,\mathrm{\,s\,}}``
`` b=\left[\frac{\,\mathrm{\,\varphi \,}}{t}\right]=\,\mathrm{\,Volt\,}``
`` c=\left[\,\mathrm{\,\varphi \,}\right]=\,\mathrm{\,Weber\,}`` (b) The emf is written as:
`` E=\frac{d\varphi }{dt}`` = 2at + b = 2 × 0.2 × 2 + 0.4 (∵ a = 0.2, b = 0.4 and c = 0.6)
On substituting t = 2 s, we get
E = 0.8 + 0.4
= 1.2 V
Page No 306:

#2-b
If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s.
Ans : The emf is written as:
`` E=\frac{d\varphi }{dt}`` = 2at + b = 2 × 0.2 × 2 + 0.4 (∵ a = 0.2, b = 0.4 and c = 0.6)
On substituting t = 2 s, we get
E = 0.8 + 0.4
= 1.2 V
Page No 306: