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Electromagnetic Induction — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction4 Marks
Question
When a current/ flows through a coil, flux linked with it is $\phi=\text{LI},$ where L is a constant known as self inductance of the coil. Any change in current sets up an induced emf in the coil. Tims, self inductance of a coil is the induced emf set up in it when the current passing through it changes at the unit rate. It is a measure of the opposition to the growth or the decay of current flowing through the coil. Also, value of self inductance depends on the number of turns in the solenoid, its area of cross-section and the relative permeability of its core material.

  1. The inductance in a coil plays the same role as:
  1. Inertia in mechanics.
  2. Energy in mechanics.
  3. Momentum in mechanics.
  4. Force in mechanics.
  1. A current of 2.5A flows through a coil of inductance 5H. The magnetic flux linked with the coil is:
  1. 0.5Wb
  2. 12.5Wb
  3. Zero
  4. 2Wb
  1. The inductance L of a solenoid depends upon its radius R as:
  1. $\text{L}\propto\text{R}$

  2. $\text{L}\propto\frac{1}{\text{R}}$

  3. $\text{L}\propto\text{R}^2$

  4. $\text{L}\propto\text{R}^3$

  1. The unit of self-inductance is:
  1. Weber ampere
  2. Weber-1 ampere
  3. Ohm second
  4. Farad
  1. The induced e.m.f. in a coil of 10 henry inductance in which current varies from 9A to 4A in 0.2 second is:
  1. 200V
  2. 250V
  3. 300V
  4. 500V 
✓

Answer

  1. (a) Inertia in mechanics.

Explanation:

The inductance in a coil plays the same role as inertia in mechanics.

  1. (b) 12.5Wb

Explanation:

Here, $\text{I}=2.5\text{A},\ \text{L}=5\text{H}$

Magnetic flux linked with the coil is,

$\phi_\text{B}=\text{LI}=(5\text{H})(2.5\text{A})=12.5\text{Wb}$

  1. (b) $\text{L}\propto\frac{1}{\text{R}}$

Explanation:

The inductance of a solenoid is $\text{L}=\mu_0{\text{n}}^2\text{Al}$ where A is the area of cross-section of the solenoid, l its length and n is the number of turns per unit length.

As, $\text{A}=\pi\text{R}^2$ where R is the radius of the solenoid.

$\therefore\text{L}=\mu_0\text{n}^2\pi\text{R}^2\text{l}\Rightarrow\text{L}\propto\text{R}^2$

  1. (c) Ohm second

Explanation:

The magnitude of induced emf is $|\varepsilon|=\text{L}\frac{\text{dl}}{\text{dt}}\Rightarrow\text{L}=\frac{|\varepsilon|\text{dt}}{\text{dl}}$

or, $\text{L}=\frac{\text{volt}\times\text{ second}}{\text{ampere}}=\text{ohm second}$

  1. (b) 250V

Explanation:

Here L = 10 henry I1 = 9A, I2 = 4A and $\triangle\text{t}=0.2$ second

Then induced e.m.f

$\varepsilon_1=-\text{L}\frac{\text{dl}}{\text{dt}}=-\text{L}\frac{\text{l}_2-\text{l}_1}{\triangle\text{t}}=\frac{-10\times(4-9)}{0.2}=\frac{50}{0.2}=250\text{V}$

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