Electromagnetic Induction Test

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Electromagnetic Induction Test - 64

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Question 1
1 / -0

Find the inductance of a unit length of two parallel wires, each of radius a, whose centers are at a distance d apart and carry equal currents in opposite directions. Neglect the flux within the wire:

A
$$\dfrac {\mu_0}{2\pi} ln \left (\dfrac {d-a}{a}\right )$$

B
$$\dfrac {\mu_0}{\pi} ln \left (\dfrac {d-a}{a}\right )$$

C
$$\dfrac {3\mu_0}{\pi} ln \left (\dfrac {d-a}{a}\right )$$

D
$$\dfrac {\mu_0}{3\pi} ln \left (\dfrac {d-a}{a}\right )$$

Solution

Using Ampere's law for one of the wires to obtain the magnetic field in terms of current $$I$$ and the variable radius $$r$$due to one of the wire
$$\mu_0\ I = \oint \vec{B}.\vec{dl}$$

$$= \int_0^{2\pi} Br\ d \theta$$
$$\mu_0\ I = B(2\pi r) $$
$$\implies B=\dfrac{\mu_0 I}{2\pi r}$$

Note that because of the symmetry of the setup, the total magnetic flux $$(\phi)$$ will be two times generated by one of the wires. Then, we continue to use the relation between self inductance and flux

$$\phi_{total}= 2 \int_A \vec{B}.\vec{dA}$$
$$=2 \int_a ^{d-a} \bigg(\dfrac{\mu_0 I}{2\pi r}\bigg ) l\ dr$$

$$\phi_{total}= \dfrac{\mu_0 Il}{\pi} ln\bigg(\dfrac{d-a}{a}\bigg)$$

Inductance, $$L= \dfrac{\phi_{total}}I= \dfrac{\mu_0 l}{\pi} ln\bigg(\dfrac{d-a}{a}\bigg)$$

Inductance per unit length , $$\phi_{total}= \dfrac{\mu_0}{\pi} ln\bigg(\dfrac{d-a}{a}\bigg)$$

Hence the correct option is $$(B)$$
Question 2
1 / -0

A rod AB moves with a uniform velocity $$v$$ in a uniform magnetic field as shown in figure.

A
The rod becomes electrically charged

B
The end A becomes positively charged

C
The end B become positively charged

D
The rod becomes hot because of Joule heating

Solution

Magnetic force $$F= q(\vec{v}\times \vec{B})= (-e)(v\hat{i} \times (-B)\hat{k})=-ebB\hat{j}$$ where we have taken the charge of electron to be (-e) where e is positive. The direction of the magnetic field is taken to be $$-B\hat{k}$$ i.e into the plane of the paper and velocity is taken along the +ve x-axis.
Thus electrons will move towards the end B and accumulate at end B.

Thus, the end A will become deficient in electrons and will be positively charged.
Question 3
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A closed loop of cross-sectional area $$10^{-2}m^2$$ which has inductance $$L=10 mH$$ and negligible resistance is placed in a time-varying magnetic field. Figure shows the variation of B with time for the interval of 4 s. The field is perpendicular to the plane of the loop (given at $$t=0, B=0, I=0)$$. The value of the maximum current induced in the loop is :

A
0.1 mA

B
10 mA

C
100 mA

D
Data insufficient

Solution

Induced e.m.f. $$ |e|=L\dfrac {dI}{dt}=A\dfrac {dB}{dt} $$
$$ \Rightarrow \int_0^I dI=\int_0^B\dfrac {A}{L}dB\Rightarrow I=\dfrac {A}{L}B $$
$$ \Rightarrow I_{max}=\dfrac {A}{L}B_{max}=\dfrac {10^{-2}}{10\times 10^{-3}}\times 0.1=0.1A=100 mA $$
Question 4
1 / -0

A solenoid has $$2000$$ turns wound over a length of $$0.3 m$$. Its cross-sectional area is equal to $$1.2\times 10^{-3} m^2$$. Around its central cross-section, a coil of 300 turns is wound. If an initial current of $$2 A$$ flowing in the solenoid is reversed in $$ 0.25 s$$, the emf induced in the coil is

A
$$0.6 mV$$

B
$$60 mV$$

C
$$40.2 mV$$

D
$$0.48 mV$$

Solution

$$emf=-\dfrac { \mu { N }_{ 1 }{ N }_{ 2 } }{ l } \dfrac { \Delta I }{ \Delta t }$$
$$ \mu =4\pi \times { 10 }^{ -7 }$$
$$ { N }_{ 1 } =2000$$
$$ { N }_{ 2 }=300$$
$$l=0.3m$$
$$ \dfrac { \Delta I }{ \Delta t } =\dfrac { 4 }{ 0.25 } A/s$$
$$\Rightarrow \left| emf \right| = \dfrac {4\pi \times 10^{-7} \times 2000 \times300 \times 4}{ 0.3 \times0.25} =$$ $$40.212mV$$
Question 5
1 / -0

The length of a thin wire require to manufacture a solenoid of length $$l=100$$ cm and inductance $$L=1 mH$$, if the solenoid's cross-sectional diameter is considerably less than its length is:

A
$$1\ km$$

B
$$0.10\ km$$

C
$$0.010\ km$$

D
$$10\ km$$

Solution

Let $$l_1$$ is the length of wire,
$$ l_1=2\pi rN\Rightarrow Nr=\dfrac {l_1}{2\pi}\Rightarrow L=\dfrac{\mu_0\pi r^2N^2}{l}=\dfrac {\mu_0\pi}{l}\dfrac {l_1^2}{4\pi^2} $$
$$ \Rightarrow l_1=\sqrt {\dfrac {Ll4\pi}{\mu_0}}=\sqrt {\dfrac {10^{-3}\times 1\times 4\pi}{4\pi\times 10^{-7}}}=10^2\ m=0.10 \ km $$
Question 6
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In Fig, there is a conducting loop ABCDEF of resistance $$\lambda$$ per unit length placed near a long straight current-carrying wire. The dimensions are shown in the figure. The lone wire lies in the plane of the loop. The current in the long wire varies as $$I=I_0(t)$$. The mutual inductance of the pair is

A
$$\dfrac {\mu_0a}{2\pi}\ln\left (\dfrac {2a+l}{l}\right )$$

B
$$\dfrac {\mu_0a}{2\pi}\ln\left (\dfrac {2a-l}{l}\right )$$

C
$$\dfrac {2\mu_0a}{2\pi}\ln\left (\dfrac {a+l}{l}\right )$$

D
$$\dfrac {\mu_0a}{2\pi}\ln\left (\dfrac {a+l}{l}\right )$$

Solution

Magnetic field at a point distance "$$x$$" from the wire, $$ B = \dfrac{ \mu I }{ 2 \pi x } $$
Total magnetic flux linking the loop, $$ \phi = \int_{l}^{l+a}{ B a. dx} + \int_{l+a}^{l+2a} { B a .dx} = \int_{l}^{l+2a}{ B a. dx}= \int_{l}^{l+2a}{ \dfrac{ \mu I }{ 2 \pi x } a .dx}=\dfrac{ \mu I a}{ 2 \pi } \ln(\dfrac{l+2a}{l})$$
Inductance of the circuit $$= L = \dfrac{flux }{I} =\dfrac{ \mu a}{ 2 \pi } \ln(\dfrac{l+2a}{l}) $$
Question 7
1 / -0

The current i in a coil varies with time as shown in the figure. The variation of induced emf with time would be :

A

B

C

D

Solution

We know, $$e = -L \dfrac {di}{dt}$$
During, $$ 0$$ to $$T/4:\ \dfrac {di}{dt} = $$ constant and $$e$$ is negative
$$T/4$$ to $$T/2$$: $$\dfrac {di}{dt} = 0$$ and $$e$$ is zero
$$T/2$$ to $$3T/4$$ : $$\dfrac {di}{dt} =$$ constant and $$e$$ is positive
Question 8
1 / -0

A long solenoid having $$200$$ turns per cm carries a current of $$1.5 amp.$$ At the centre of it is placed a coil of $$100$$ turns of cross-sectional are $$3.14\times 10^{-4}m^2$$ having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within $$0.05 sec,$$ the induced e.m.f. in the coil is

A
$$0.48 V$$

B
$$0.048 V$$

C
$$0.0048 V$$

D
$$48 V$$

Solution

$$B=\mu_0ni=(4\pi \times 10^{-7})(200\times 10^{-2})\times 1.5$$

$$=3.8\times 10^{-2}Wb/m^2$$

Magnetic flux through each turn of the coil

$$\phi=BA=(3.8\times 10^{-2})(3.14\times 10^{-4})=1.2\times 10^{-5}weber$$

When the current in the solenoid is reversed, the change in magnetic flux

$$=2\times (1.2\times 10^{-5})=2.4\times 10^{-5}weber$$

Induced e.m.f. $$=N\dfrac {d\phi}{dt}=100\times \dfrac {2.4\times 10^{-5}}{0.05}=0.048V$$.
Question 9
1 / -0

A wire shaped as a circle of radius R rotates about the axis OO' with an angular velocity $$\omega$$ as shown in figure. Resistance of the circuit is $$R$$. Find the mean thermal power generated in the loop during a period of a rotation.

A
$$\dfrac{(B \pi a^2 \omega)^2}{4R}$$

B
$$\dfrac{(B \pi a^2 \omega)^2}{2R}$$

C
$$\dfrac{(3B \pi a^2 \omega)^2}{2R}$$

D
None of these

Solution

$$A = \pi a^2 \cos \omega t; \phi = BA = B \pi a^2 \cos \omega t$$
$$\displaystyle i = \dfrac{\varepsilon}{R} = - \dfrac{d\phi}{dt} /R = \dfrac{B \pi a^2 \omega \sin \omega t}{R}$$ and
$$<P>= \displaystyle \dfrac{1}{T} \int_0^T i^2 Rdt = \dfrac{(B \pi a^2 \omega)^2}{2R}$$
Question 10
1 / -0

A rectangular loop with a slide wire of length $$l$$ is kept in a uniform magnetic field as shown in figure (a). The resistance of slider is $$R$$. Neglecting self inductance of the loop find the current in the connector during its motion with a velocity $$v$$.

A
$$\displaystyle \dfrac{Blv}{R_1 + R_2 + R}$$

B
$$\displaystyle \dfrac{Blv(R_1+ R_2)}{R(R_1 + R_2)}$$

C
$$\displaystyle \dfrac{Blv(R_1 + R_2)}{RR_1 + RR_2 + R_1R_2}$$

D
$$\displaystyle Blv \left ( \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right )$$

Solution

The equivalent circuit is shown in figure (c).
Obviously, $$I = \displaystyle \dfrac{Blv}{R + \dfrac{R_1R_2}{R_1 +R_2}} = \dfrac{Blv(R_1 + R_2)}{RR_1 + RR_2 + R_1 R_2}$$