Electromagnetic Induction Test

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Electromagnetic Induction Test - 27

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Weekly Quiz Competition

Question 1
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A solenoid of inductance $$L$$ and length $$l$$ and mass $$m$$, whose windings are made of material of density $$D$$ and resistivity is $$\rho$$,(the winding resistance is $$R$$). Find out the inductance in terms of others.

A
$$\dfrac {\mu_0}{4\pi l}\dfrac {Rm}{\rho D}$$

B
$$\dfrac {\mu_0}{4\pi R}\dfrac {lm}{\rho D}$$

C
$$\dfrac {\mu_0}{4\pi l}\dfrac {R^2m}{\rho D}$$

D
$$\dfrac {\mu_0}{2\pi R}\dfrac {lm}{\rho D}$$

Solution

For a solenoid, $$L=\mu_0 N^2\dfrac {A}{l}$$. If $$x$$ is the length of the wire and $$a$$ is the area of cross-section, then

Resistance $$R=\dfrac {\rho x}{a}$$ and mass $$m=axD$$

$$Rm=\dfrac {\rho x}{a}axD, x=\sqrt {\dfrac {Rm}{\rho D}}$$

Also, $$x=2\pi rN, N=\dfrac {x}{2\pi r} \left (\therefore L=\dfrac {\mu_0N^2A}{l}\right )$$

$$\therefore L=\mu_0\left (\dfrac {x}{2\pi r}\right )^2\dfrac {\pi r^2}{l}=\dfrac {\mu_0}{4\pi l}\dfrac {Rm}{\rho D}$$
Question 2
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The self induction takes place when magnetic flux through a coil :

A
Remains steady

B
Decreases

C
Increases

D
Either (B) or (C)

Solution

As per Faradays law, EMF is always directly proportional to the rate of change of magnetic flux. Therefore, electromagnetic induction takes place whenever the magnetic flux passing through the object changes.
Question 3
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A toroid is wound over a circular core with total no. of turns equal to $$N$$. The radius of each turn is $$r$$ and radius of toroid is $$R (> > r)$$. The coefficient of self-inductance of the toroid is given by

A
$$L=\dfrac {\mu_0Nr^2}{2R}$$

B
$$L=\dfrac {\mu_0Nr}{2R}$$

C
$$L=\dfrac {\mu_0Nr^2}{R}$$

D
$$L=\dfrac {\mu_0N^2r^2}{2R}$$

Solution

$$L=\dfrac {\phi}{I}, \phi=NAB$$
$$B=\mu_0nI$$
where $$n=\dfrac {N}{2\pi R}$$
$$\therefore \phi=N\pi r^2\left (\mu_0\dfrac {N}{2\pi R}I\right )$$
$$\phi=\dfrac {\mu_0N^2r^2I}{2R}$$
$$L=\dfrac {\phi}{I}=\dfrac {\mu_0N^2r^2}{2R}$$
Question 4
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A wire of fixed length is wound on a solenoid of length $$l$$ and radius $$r$$. Its self-inductance is found to be $$L$$. Now, if the same wire is wound on a solenoid of length $$l/2$$ and radius $$r/2$$, then the self-inductance will be

A
$$2L$$

B
$$L$$

C
$$4L$$

D
$$8L$$

Solution

$$L=\dfrac {\mu_0N^2\pi r^2}{l}$$
Length of wire $$=N2\pi r= Constant (=C$$, suppose)
$$\therefore L=\mu_0\left (\dfrac {C}{2\pi r}\right )^2\dfrac {\pi r^2}{l}$$
$$\therefore L\propto \dfrac {1}{l}$$
$$\therefore$$ Self-inductance will become $$2L$$.
Question 5
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The length of a wire required to manufacture a solenoid of length $$l$$ and self-induction $$L$$ is (cross-sectional area is negligible)

A
$$\sqrt {\dfrac {2\pi Ll}{\mu_0}}$$

B
$$\sqrt {\dfrac {\mu_0Ll}{4\pi}}$$

C
$$\sqrt {\dfrac {4\pi Ll}{\mu_0}}$$

D
$$\sqrt {\dfrac {\mu_0Ll}{2\pi}}$$
Solution
Hint: Use the formula of self-inductance of a solenoid.

Correction Option: C

Explanation for correct option:
$$\textbf{Step1: Find number of turns in the solenoid and area of cross section of solenoid}$$
- Let $$x$$ is the length of the wire, then the length of the solenoid is $$x = 2\pi rN$$. (Here N$$\Rightarrow$$ number of turns}
$$ \Rightarrow N = \dfrac{x}{{2\pi r}}$$

Since, the wire is in the shape of a cylinder, the area of cross section is $$A = \pi {r^2}$$ (Here $$r$$ is radius of coil)
$$\textbf{Step2: Find length of a wire required}$$
- The self-inductance of a solenoid is given by, $$L = \dfrac{{{\mu _0}{N^2}A}}{l}$$, where $${\mu _0}$$ is the magnetic permeability, $$l$$ is the length of the solenoid and $$A$$ is the area of each turn in the solenoid.
$$ \Rightarrow L = \dfrac{{{\mu _0}{{\left( {\dfrac{x}{{2\pi r}}} \right)}^2}\pi {r^2}}}{l}$$

$$ \Rightarrow x = \sqrt {\dfrac{{4\pi Ll}}{{{\mu _0}}}} $$

So, option C is correct.
Question 6
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Two coils X and Y are linked such that emf $$E$$ is induced in Y when the current in X is changing at the rate $$I'(=dI/dt)$$. If a current $$I_0$$ is now made to flow through Y, the flux linked with X will be

A
$$EI_0I'$$

B
$$\dfrac {I_0I'}{E}$$

C
$$\left (\dfrac {E}{I_0}\right )I'$$

D
$$\left (\dfrac {E}{I'}\right )I_0$$

Solution

Let $$ M $$ be the mutual inductance,
We can write $$ E=MI' $$
We know that $$ E=-\dfrac{d\Phi_B}{dt} $$
$$ \therefore \Phi_B=-\int Edt=-M\int I'dt=-MI $$ where$$ I $$ is the current in the circuit.
Given $$ I=I_0 $$.
So $$ \Phi_B=-MI_0=-(\dfrac{E}{I'})I_0 $$.
Sign is indicative of the direction. So for magnitude only
$$ \Phi_B=(\dfrac{E}{I'})I_0 $$
Question 7
1 / -0

The time constant of an inductance coil is $$2\times 10^{-3}s$$. When a $$90\Omega$$ resistance is joined in series, the same constant becomes $$0.5\times 10^{-3}s$$. The inductance and resistance of the coil are

A
$$30 mH; 30\Omega$$

B
$$60 mH; 30\Omega$$

C
$$30 mH; 60\Omega$$

D
$$60 mH; 60\Omega$$

Solution

Initally the time constant can be representyed as:
$$\dfrac{L}{R}=2\times 10^{-3}$$..................(i)

When the resistance is increased, the new time constant is given as:
$$\dfrac{L}{R+90}=0.5 \times 10^{-5}$$..................(ii)

From (i) and (ii) , on solving , we get
$$L=60 mH$$ and $$R= 30 \Omega$$
Question 8
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Two coils are at fixed locations. When coil 1 has no current and the current in coil 2 increases at the rate of $$15.0 As^{-1}$$, the emf in coil 1 is $$25 mV$$. When coil 2 has no current and coil 1 has a current of $$3.6A$$, the flux linkage in coil 2 is

A
$$16 mWb$$

B
$$10 mWb$$

C
$$4.00 mWb$$

D
$$6.00 mWb$$

Solution

$$ |M|=\dfrac {e_1}{(di_2/dt)}=\dfrac {\phi_2}{i_1} $$
$$ \therefore \phi_2=\dfrac {e_1i_1}{(di_2/dt)}=\dfrac {(25.0\times 10^{-3})(3.6)}{(15)} $$
$$ =6\times 10^{-3}=6 mWb $$
Question 9
1 / -0

Calculate the inductance of a unit length of a double tape line as shown in figure. The tapes are separated by a distance $$h$$ which is considerably less than their width $$b$$.

A
$$\dfrac {\mu_0h}{b}$$

B
$$\dfrac {\mu_0h}{2b}$$

C
$$\dfrac {2\mu_0h}{b}$$

D
$$\dfrac {\sqrt 2\mu_0h}{b}$$

Solution

$$\oint \vec B\cdot d\vec l=\mu_0I\Rightarrow B.2b=\mu_0I$$
$$\Rightarrow B=\dfrac {\mu_0I}{2b}\rightarrow$$ due to one tape
Net field $$=2B$$
Magnetic flux passing through this double tape
$$\phi =2BA=2B(lh)\Rightarrow \phi=\dfrac {\mu_0I}{b}lh$$
$$L=\dfrac {\phi}{I}=\dfrac {\mu_0lh}{b}\Rightarrow \dfrac {L}{l}=\dfrac {\mu_0h}{b}$$
Question 10
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The coil is wound on an iron core and looped back on itself so that core has two sets of closely wound coils carrying current in opposite directions. The self inductance is

A
$$0$$

B
$$2L$$

C
$$2L + M$$

D
$$L + 2M$$

Solution

Let the inductance due to first loop alone be $$L$$.
Then inductance due to second loop alone is also $$L$$.
The effective self inductance can be calculate by: $$L_{eff}=L_1+L_2-2M$$$$=L+L-2\sqrt{L.L}=0$$

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Electromagnetic Induction Test - 27

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Electromagnetic Induction Test - 27

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Question 1

$$\dfrac {\mu_0}{4\pi l}\dfrac {Rm}{\rho D}$$

$$\dfrac {\mu_0}{4\pi R}\dfrac {lm}{\rho D}$$

$$\dfrac {\mu_0}{4\pi l}\dfrac {R^2m}{\rho D}$$

$$\dfrac {\mu_0}{2\pi R}\dfrac {lm}{\rho D}$$

Solution

Question 2

Remains steady

Decreases

Increases

Either (B) or (C)

Solution

Question 3

$$L=\dfrac {\mu_0Nr^2}{2R}$$

$$L=\dfrac {\mu_0Nr}{2R}$$

$$L=\dfrac {\mu_0Nr^2}{R}$$

$$L=\dfrac {\mu_0N^2r^2}{2R}$$

Solution

Question 4

$$2L$$

$$L$$

$$4L$$

$$8L$$

Solution

Question 5

$$\sqrt {\dfrac {2\pi Ll}{\mu_0}}$$

$$\sqrt {\dfrac {\mu_0Ll}{4\pi}}$$

$$\sqrt {\dfrac {4\pi Ll}{\mu_0}}$$

$$\sqrt {\dfrac {\mu_0Ll}{2\pi}}$$

Solution

Question 6

$$EI_0I'$$

$$\dfrac {I_0I'}{E}$$

$$\left (\dfrac {E}{I_0}\right )I'$$

$$\left (\dfrac {E}{I'}\right )I_0$$

Solution

Question 7

$$30 mH; 30\Omega$$

$$60 mH; 30\Omega$$

$$30 mH; 60\Omega$$

$$60 mH; 60\Omega$$

Solution

Question 8

$$16 mWb$$

$$10 mWb$$

$$4.00 mWb$$

$$6.00 mWb$$

Solution

Question 9

$$\dfrac {\mu_0h}{b}$$

$$\dfrac {\mu_0h}{2b}$$

$$\dfrac {2\mu_0h}{b}$$

$$\dfrac {\sqrt 2\mu_0h}{b}$$

Solution

Question 10

$$0$$

$$2L$$

$$2L + M$$

$$L + 2M$$

Solution

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