Magnetism and Matter Test 48

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Magnetism and Matter Test 48

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Question 1
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Charge Q is uniformly distributed on a thin insulating ring of mass m which is initially at rest. To what angular velocity will the ring be accelerated when a magnetic field B, perpendicular to the plane of the ring, is switched on?

A
$$\dfrac {QB}{2m}$$

B
$$\dfrac {3QB}{2m}$$

C
$$\dfrac {QB}{m}$$

D
$$\dfrac {QB}{4m}$$

Solution

In accordance with Faraday's law of electromagnetic induction, the changing magnetic field induces an electric field in the ring. Let us imagine the ring to be divided into differential elements of length ds and denote the tangential component of the induced electric field by $$E_1$$. The charge on element ds of the ring is $$dQ=Q\dfrac {ds}{2\pi r}$$,
Where r is the radius of the ring.The force exerted on it is $$dF_t=dQE_t$$, and the resultant torque is $$d \tau=rdf_t$$.

Thus the total toque experienced by the ring is
$$\tau=\int d \tau= \int rQ \dfrac {ds}{2 \pi r} E_t= \dfrac {Q}{2 \pi}\int E_1 ds$$

The induced electromotive force along the ring is directly proportional to the rate of change in the magnetic flux, we have

$$\int E_1 ds= \dfrac {d\phi}{dt}==\pi r^2 \dfrac {dB}{dt}$$

As a result of the toque, the ring, which has a moment of inertia $$I=mr^2$$ , starts to spin with angular acceleration $$\alpha$$. During a time interval dt its angular velocity changes by

$$d\omega=\alpha dt=\dfrac {\tau}{1}dt= \dfrac {Q}{2 \pi}(\pi r^2 \dfrac {dB}{dt}) \dfrac {1}{mr^2} dt=\dfrac {Q}{2m}dB$$

Since the magnetic field strength increases from zero to B, The final angular velocity of the ring will be

$$\omega=(QB/2m)$$
$$\Rightarrow$$ The final velocity does not depend on the radius of the ring, the time over which the magnetic flux changes, or even on how the magnetic flux increase with time.

$$\Rightarrow$$ In our calculation we ignored the magnetic field produced by the rotating ring.

$$\Rightarrow$$Except in the case of a cylindrical symmetric uniform field, it is not possible to find the actual volume of the induced electric field within the ring because the geometrical structure of the magnetic field is unknown and we do not know the position of the ring in the magnetic field. We can determine the total induced electromotive force, but not the electric field itself.
Question 2
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A standing wave $$y=2A \sin kx \cos \omega t$$ is set up in the wire AB fixed at both the ends by two vertical walls . The region between the walls contains a constant magnetic field B. Now, answer the following question.

The wire is found to vibrate in the third harmonic. The maximum emf induced is

A
$$ \dfrac {4(AB) \omega}{k}$$

B
$$ \dfrac {3(AB) \omega}{k}$$

C
$$ \dfrac {2(AB) \omega}{k}$$

D
$$ \dfrac {(AB) \omega}{k}$$

Solution

Given
$$y= 2A sin \,k\,x \,cos \omega t$$

on differentiating we get,

$$v= \dfrac {dy}{dt}=-2A sin kx \omega sin \omega t $$

$$v_{max}=-2A sin kx , \ k= \dfrac {3 \pi}{AB}$$

now,
$$e= \int _0^{t=AB} Bv_{max} dx$$

$$= -2 A \omega B \int _0^{AB} sin kx dx$$

$$= + \dfrac {2\omega AB}{k}[cos \dfrac {3 \pi}{AB} -cos \theta]$$

$$=-\dfrac {4(AB) \omega}{k}$$
Question 3
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A standing wave $$y=2A sin kx cos \omega t$$ is set up in the wire AB fixed at both the ends by two vertical walls . The region between the walls contains a constant magnetic field B. Now, answer the following question.

In which of the following modes the emf induced in AB is always zero?

A
fundamental node

B
second harmonic

C
second overtone

D
fourth overtone

Solution
Question 4
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Consider two parallel, conducting frictionless tracks kept in a gravity-free space as shown in fig. A movable conductor PQ initially kept at OA, is given a velocity $$10 m s^{-1}$$ towards the right. Space contains a magnetic field which depends upon the distance moved by conductor PQ from the OA line and given by $$\vec { B } =cx (-\hat {k}) $$[c=constant =1 S.I unit] the mass of the conductor PQ is 1 kg length of PQ is 1 m. Answer the following question based on the above passage.

The distance traveled by the conductor when its speed is 5 m/s is

A
$$(\dfrac {15}{2})^{1/3}$$

B
$$(\dfrac {10}{3})^{1/3}$$

C
$$10^{1/3}$$

D
none of these

Solution

Consider anticlockwise direction is positive

$$\phi= B A cos 180^{\circ}=cl x/x \, cos 180^{\circ}=-clx^2$$

$$e= -\dfrac {d\phi}{dt}=cl2x \dfrac {dx}{dt}$$,

$$I=\dfrac {e}{R}=\dfrac {2clxv}{1}=2clxv$$

This is positive so the current is anticlockwise.

$$F= m(-a) \Rightarrow IBl=-mv \dfrac {dv}{dx}$$

$$ \Rightarrow 2c^2l^2x^2v=-mv \dfrac {dv}{dx}$$

on integrating,

$$ \Rightarrow 2c^2 l^2 \int_0^x x^2dx=-\int_{10}^5 mdv $$

$$\Rightarrow 2 C^2l^2 \dfrac {x^3}{3}= m(5)$$

$$\Rightarrow x^3= \dfrac {15m}{2c^2l^2}$$

$$ \Rightarrow x=(\dfrac {15}{2})^{1/3}$$
Question 5
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Consider two parallel, conducting frictionless tracks kept in a gravity-free space as shown in fig. A movable conductor PQ, initially kept at OA, is given a velocity $$10 m s^{-1}$$ towards right. The space contains a magnetic field which depends upon the distance moved by conductor PQ from the OA line and given by $$\vec { B } =cx (-\hat {k}) $$[c=constant =1 S.I unit] the mass of the conductor PQ is 1 kg length of PQ is 1 m. Answer the following question based on the above passage.

The work done by magnetic force acting on the conductor PQ during its motion in the time interval t=0 to t=t seconds when the speed of conductor is $$5 m\, s^{-1} $$ is

A
zero

B
50 J

C
10 J

D
30 J

Solution

As $$\vec F=q(\vec v\times \vec B)$$ ie,

$$\vec F \ is\ perpendicular \ to\ \vec V$$ ,

workdone is zero

Magnetic force is not doing any work.
Question 6
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A metal rod of resistance $$ 20 \Omega $$ is fixed along a diameter of a conducting ring of radius 0.1 m and lies on x-y plane. There is a magnetic field $$ \overrightarrow {B }=(50T) \hat k$$. the ring rotated with angular velocity $$\omega=20 rad/s$$ about its axis. An external resistance of $$ 10 \Omega $$ is connected across the centre of the ring and rim. The current through external resistance is

A
$$\dfrac {1}{4}$$

B
$$\dfrac {1}{2}$$

C
$$\dfrac {1}{3}$$

D
0

Solution

From the figure
emf induced between the centre of the ring and the rim is

$$e= \dfrac {1}{2} B \omega R^2=\dfrac {1}{2}(50(20)(0.1)^2=5V$$
the circuit reduces to the second figure

$$\therefore i= \dfrac {5}{10+5}=\dfrac {1}{3}A$$
Question 7
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The radius of the circular conducting loop shown in fig. is R. Magnetic field is decreasing at a constant rate $$\alpha$$ Resistance per unit length of the loop is $$\rho$$then the current in wire AB is (AB is one of the diameter)

A
$$\dfrac {R \alpha}{2\rho}$$ from A to B

B
$$\dfrac {R \alpha}{2\rho}$$ from B to A

C
$$\dfrac {R \alpha}{\rho}$$ from A to B

D
0

Solution
Question 8
1 / -0

A solenoid has an inductance of $$60$$ henrys and a resistance of $$30$$ ohms. If it is connected to a $$100$$ volt battery, how long will it take for the current to reach $$\dfrac{e - 1}{e} \approx 63.2\%$$ of its final value

A
$$1$$ second

B
$$2$$ seconds

C
$$e$$ seconds

D
$$2e$$ seconds

Solution
Question 9
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A small coil C with $$ N = 200 $$ turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The cross sectional area of coil is $$A= 1.0 cm$$ length of arm OA of the balance beam is $$I=30 cm.$$
When there is no current in the coil the balance is in equilibrium. On passing a current $$ I = 22 mA $$ through the coil the equilibrium is restored by putting the additional counter weight of mass $$ \theta m = 60 mg $$ on the balance pan. Find the magnetic induction at the spot where coil is located.

A
$$ 0.4 T $$

B
$$ 0.3 T $$

C
$$ 0.2 T $$

D
$$ 0.1 T $$

Solution

On passing current through the coil, it acts as a magnetic dipole. Torque acting on magnetic dipole is counter balanced by the moment of additional weight about position O. Torque acting on a magnetic dipole.
$$ \tau = MB sin \theta = (Ni A )B sin 90^0 = NiAB $$
Again $$ \tau =Force \times lever arm = \Delta mg \times l $$
$$ \Rightarrow NiAb = \Delta mgl $$
$$ \Rightarrow B = \dfrac { \Delta mgl}{NiA} = \dfrac { 60 \times 10^{-3} \times 9.8 \times 30 \times 10^{-2} }{ 200 \times 22 \times 10^{-3} \times 1 \times 10^{-4} } = 0.4 T $$
Question 10
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The magnetic field in the cylindrical region shown in figure increases at a constant rate of $$20\, mT/sec.$$ Each side of the square loop $$ABCD$$ has a length of $$1 \,cm$$ and resistance of $$4\Omega$$. Find the current in the wire AB if the switch S is closed

A
$$1.25 \times 10^{-7} \,A,$$ (anti-clockwise)

B
$$1.25 \times 10^{-7} \,A$$ (clockwise)

C
$$2.5 \times 10^{-7} \,A$$ (anti clockwise)

D
$$2.5 \times 10^{-7} \,A$$ (clockwise)

Solution

Given, Magnetic field in the cylindrical region increases at a constant rate of $$20mT/sec$$. Each side of square loop has length $$1$$ cm and resistor $$4\Omega$$.
To find, current in the wire AB when switch is closed.
We know,

$$e=n\Delta \Phi$$

$$\Rightarrow e=\dfrac{d\Phi}{dt}$$ and $$\Phi =BA$$

$$\Rightarrow e=\dfrac{dB}{dt}(1\times 10^{-4})$$

and $$e=IR$$

$$\Rightarrow \dfrac{dB}{dt}=IR$$

As each side has resistance R

$$\therefore R_{eq}=4+4+4+4$$

$$\Rightarrow R_{eq}=16\Omega$$

Now, $$\dfrac{dB}{dt}\times (1\times 10^{-4})=I\times 16$$

$$\Rightarrow 20\times 10^{-3}\times 10^{-4}=I\times 16$$

$$\Rightarrow I =\dfrac{20\times 10^{-7}}{16}$$

$$\Rightarrow I=1.25\times 10^{-7}A$$
By using Lenz's Law, Direction of current will be Anticlockwise.
Therefore, option A correct.