Magnetism and Matter Test 40

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Magnetism and Matter Test 40

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Question 1
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A bar magnet of magnetic moment $$M$$ is divided into '$$n$$' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength $$2T$$ and held making an angle $$60^{0}$$ with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:

A
$$\dfrac{M}{n}$$J

B
$$Mn$$ J

C
$$\dfrac{M}{n^{2}}$$J

D
$$Mn^{2}$$ J

Solution

Answer is A
Net magnetic moment after cutting$$= \dfrac{M}{n}$$
We know that, $$ |U|= MB \cos \theta$$
In equilibrium position, all the potential energy will be converted into kinetic energy i.e. $$ KE=|U| $$
Given $$ M_{new}=\dfrac{M}{n}, B=2T, \theta=60^{\circ} $$
Therefore, $$K.E= \dfrac{M}n\times 2\times \cos 60^{\circ}=\dfrac{M}{n} J $$
Question 2
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Assertion : Lenz's Law violates the principle of conservation of energy
Reason : Induced emf always opposes the change in magnetic flux responsible for its production.

A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

C
Assertion is true but Reason is false

D
Assertion is false but Reason is true

Solution

A: Lenz does not violate it. It is based on it.
R: Lenz law.
Question 3
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The least value of magnetic moment (where $$m$$ is the mass of the electron, $$e$$ is the charge of electron) is :

A
$$\dfrac{eh}{m}$$

B
$$\dfrac{eh}{4\pi m}$$

C
$$\dfrac{2eh}{\pi m}$$

D
$$\dfrac{eh}{\pi m}$$

Solution

For an electron moving in an orbit around a nucleus produces an average current along its orbit.
$$\therefore $$ Period of motion, $$T=\cfrac { distance }{ velocity } =\cfrac { 2\pi r }{ V } $$
Magnetic moment associated = current X area
$$\mu =\cfrac { charge }{ time } \times area$$
$$\mu =\cfrac { eV }{ 2\pi r } \times \pi { r }^{ 2 }=\cfrac { eVr }{ 2 } $$
Angular momentum, $$L=mvr=\cfrac { nh }{ 2\pi } $$
$$\therefore vr=\cfrac { nh }{ 2\pi m } $$
where $$n$$ is quantum number associated.
$$\therefore \mu =\cfrac { e }{ 2 } \left( \cfrac { nh }{ 2\pi m } \right) $$
$$\therefore { \mu }_{ min. }=\cfrac { eh }{ 4\pi m } $$ (at $$n=1$$)
Question 4
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The given figure shows planar loops of different shapes moving out of or into a region of a magnetic field, which is directed normal to the plane of the loop away from the reader. Then the direction of induced current in each loop using Lenz's law will be

a). The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field. The induced current must flow along the path bcdab so that it opposes the increasing flux.

b) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb,
so as to oppose the change in flux.

c) As the magnetic flux decreases, due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows
along cdabc, so as to oppose change in flux.

Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field.

The correct statements are

A
a,b

B
a,c

C
b,c

D
a,b,c

Solution

The flux associated with the rectangle $$ abcd $$ is increasing while that with the triangle $$ abc $$ and loop $$ abcd $$ is decreasing.
To check whether the direction of current given in the options produces magnetic field opposing this change i.e. in the supposed direction, we can use the right hand grip rule.
Question 5
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Directions For Questions

In the figure shown the loop placed in the plane of the paper has radius '$$R$$', self inductance '$$L$$' and zero resistance. A uniform magnetic field $$B=B_0 \sin \omega t$$ is applied ($$B_0$$ and $$\omega$$ are positive constants) normal to the plane of the paper. Initially the loop has zero current.

...view full instructions

Direction of current in the loop at $$t=\dfrac{\pi}{3 \omega}$$ :

A
clockwise

B
anticlockwise

C
no current will flow at that time

D
cannot be determined

Solution

$$B = B_o sin(wt)$$
$$\cfrac{d B}{dt} = B_ow cos(wt) $$
at $$t = \cfrac{\pi}{3 \omega}$$
$$\cfrac{d B}{dt} = B_ow/2 $$
The magnetic field is increasing.
According to Lenz's law, the current is induced to resist the change in magnetic field.
Hence According to right hand law current is in anticlockwise direction.
Question 6
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Two coils carrying current in opposite direction are placed co-axially with centres at some finite separation. If they are brought close to each other then current flowing in them should

A
decrease

B
increase

C
remain same

D
become zero

Solution

As shown in figure,
according to right hand rule, the direction of magnetic field is opposite in both the wires.
When brought closer induced effects should produce repulsion. So currents should increase, so that pole strength increases. Hence repulsion increases
Question 7
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A small circular loop is suspended from a insulating thread. Another co-axial circular loop carrying a current $$I$$ and having radius much larger than the first loop starts moving towards the smaller loop. The smaller loop will :

A
be attracted towards the bigger loop

B
be repelled by the bigger loop

C
experience no force

D
show all of the above

Solution

Lenz's Law
If an induced current flows, its direction is always such that it will oppose the change which produced it.
So, when the first loop is moved towards the smaller loop it will face repulsion.
(i.e. opposition to the change being caused)
Question 8
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The ring $$B$$ is coaxial with a solenoid $$A$$ as shown in figure. As the switch $$S$$ is closed at $$t=0$$, the ring $$B$$

A
is attracted towards $$A$$

B
is repelled by $$A$$

C
is initially repelled and then attracted

D
is initially attracted and then repelled

Solution

Current increases with time. So flux passing through B will increase with time. From Lenz's law, it should have a tendency to move away from the coil to decrease flux.
Question 9
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Directions For Questions

Electrical resistance of certain material, known as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature $$T_c(0)$$. An interesting property of superconductors is that their critical temperature becomes smaller than $$T_c(0)$$ if they are placed in a magnetic field, i.e, the critical temperature $$T_c(B)$$ is a function of the magnetic field strength B. The dependence of $$T_c(B)$$ on B is shown in fig.

...view full instructions

In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields $$B_1$$ (solid line) and $$B_2$$ (dashed line). If $$B_2$$ is larger than $$B_1$$, which of the following graphs shows the correct variation of R and T in these fields?

A

B

C

D

Solution

Resistance usually varies directly with temperature.
But when considering superconductors, the least temperature is Tc(0).
In presence of B, Tc decreases. But that doesn't say anything about general R and T relation.
Only info we have is Tc($$B_2$$) < Tc($$B_1$$) if $$B_2>B_1$$. (from observing the graph)
Question 10
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Two identical conductors $$P$$ and $$Q$$ are placed on two frictionless rails $$R$$ and $$S$$ in a uniform magnetic field directed into the plane. If $$P$$ is moved in the direction shown in figure with a constant speed, then rod $$Q$$.

A
will be attracted toward $$P$$

B
will be repelled away from $$P$$

C
will remain stationary

D
may be repelled away or attracted toward $$P$$

E
answer required

Solution

As the conductor P moves away from Q, the area of the loop enclosed by the conductors and the rails increases. This is turn increases the flux through the loop.
EMF will be induced in such a way that the change in flux will be resisted. The induced current will cause Q to move towards P thereby reducing the area and thus the flux back.