Magnetism and Matter Test 41

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Magnetism and Matter Test 41

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Question 1
1 / -0

Three identical coils $$A,B,C$$ carrying currents are placed coaxially with their planes parallel to one another. $$A$$ and $$C$$ carry currents as shown in figure. $$B$$ is kept fixed, while $$A$$ and $$C$$ both are moved toward $$B$$ with the same speed. Initially, $$B$$ is equally separated from $$A$$ and $$C$$. The direction of the induced current in the coil $$B$$ is :

A
same as that in coil $$A$$

B
same as that in coil $$B$$

C
zero

D
none of these

Solution

The rings can be treated as two small bar magnets with polarities as shown.
Both A and C are equidistant from initially and are moving towards B with same speed. Therefore they will always be equidistant from B.
At anytime the flux in the coil B due to coil A will be equal and opposite to the flux in the coil B due to coil C.
Thus, at any time, the net flux through coil B is zero and so is the change in flux.
So no current will be induced in coil B
Question 2
1 / -0

An electron moves on a straight line path $$YY'$$ as shown in figure. A coil is kept in the right such that $$YY'$$ is in the plane of the coil. At the instant when the electron gets closest to the coil (neglect self-induction of the coil),

A
the current in the coil flows clockwise

B
the current in the coil flows anticlockwise

C
the current in the coil is zero

D
the current in the coil does not change the direction as the electron crosses point $$O$$

E
answer required

Solution

The current through the coil is zero as there is not charge in current through the wire therefore there is no induction.
So the correct answer is C.
Question 3
1 / -0

A square coil $$ACDE$$ with it's plane vertical is released from rest in a horizontal uniform magnetic field $$\vec { B } $$ of length $$2L$$ ( as shown in figure). The acceleration of the coil is

A
less than $$g$$ for all time till the loop crosses the magnetic field completely

B
less than $$g$$ when it enters the field and greater than $$g$$ when it comes out of the field

C
$$g$$ all the time

D
less than $$g$$ when it enters and comes out of the field but equal to $$g$$ when it is within the field

Solution

Flux through the loop only changes when it enters and when it leaves the field. When it is fully inside the field or fully outside the field, the flux doesnt change.
When the flux changes, there is a current induced that opposes the cause according to Lenz's law. Therefore only when enetering and when leaving the field, the acceleration of the loop will be less than $$g$$. Other times, when it is fully inside the field or fully outside the field, the acceleration will be $$g$$.
Question 4
1 / -0

A conductor of length $$l$$ and mass $$m$$ can slide without any friction along the two vertical conductors connected at the top through a capacitor (figure). A uniform magnetic field $$B$$ is set up $$\bot $$ to the plane of paper. The acceleration of the conductor

A
is constant

B
increases

C
decreases

D
cannot be determined

Solution

using Lenz's law there will be an upward force acting on the conductor and this force increases as the velocity increases, thus, the acceleration of the conductor decreases.
Question 5
1 / -0

In the space shown a non-uniform magnetic field $$\vec { B } ={B}_{0}(1+x)(\hat {-k } )$$ tesla is present. A closed loop of small resistance, placed in the x-y plane is given velocity $${V}_{0}$$. The force due to magnetic field on the loop is

A
zero

B
along $$+x$$ direction

C
along $$-x$$ direction

D
along $$+y$$ direction

E
none of these

Solution

Field is constant along the y direction. As the loop moves towards x direction, field changes. This changes the flux and by Lenz's a law current is induced that opposes this motion along x direction.
Therefore force will be along -x direction.
Question 6
1 / -0

Two disc of radius R and 2R are made up of same insulating material both has uniform charge density $$\displaystyle \sigma $$ on one surface. Both the discs are rotated about centrodial axis with constant angular velocity $$\displaystyle \omega $$ in uniform magnetic field B. If $$\displaystyle E_{1}\: and\: E_{2} $$ are the work done by external agent to provide this energy to the disc Calculate $$\displaystyle E_{2}/2E_{1} $$ :

A
8

B
10

C
12

D
15
Question 7
1 / -0

The potential energy of a bar magnet of magnetic moment $$M$$ placed in a magnetic field of induction $$$$. The position of stable equilibrium of the magnet is at the angular position given by $$\theta$$ equal to:

A
$$0^0$$

B
$$90^0$$

C
$$45^0$$

D
$$180^0$$

Solution

For stable equilibrium magnet field $$\vec{B}$$ and magnetic moment $$'M'$$ should be in parallel then angle $$\theta=0^\circ$$.
Question 8
1 / -0

A coil of radius R carries a current I. Another concentric coil of radius $$\displaystyle r\left( r<<R \right) $$ caries current $$\displaystyle \frac { I }{ 2 } $$. Initially planes of the two coils are mutually perpendicular and both the coils are free to rotate about common diameter. They are released from rest from this position. The masses of the coils are M and m respectively $$\displaystyle \left( m<M \right) $$. During the subsequent motion let $$\displaystyle { K }_{ 1 }$$ and $$\displaystyle { K }_{ 2 }$$ be the maximum kinetic energies of the two coils respectively and let U be the magnitude of maximum potential energy of magnetic interaction of the system of the coils. Choose the correct options.

A
$$\displaystyle \frac { { K }_{ 1 } }{ { K }_{ 2 } } =\frac { M }{ m } { \left( \frac { R }{ r } \right) }^{ 2 }$$

B
$$\displaystyle { K }_{ 1 }=\frac { { Umr }^{ 2 } }{ { mr }^{ 2 }+{ MR }^{ 2 } } { K }_{ 2 }=\frac { { UMR }^{ 2 } }{ { mr }^{ 2 }+{ MR }^{ 2 } } $$

C
$$\displaystyle U=\frac { { \mu }_{ 0 }\pi { I }^{ 2 }{ r }^{ 2 } }{ 4R } $$

D
$$\displaystyle { K }_{ 2 }>>{ K }_{ 1 }$$

Solution

Coil 1 Coil 2
Radius R r(<<R)
Current I I/2
Mass M m
Kinetic energy $$K_1$$ $$K_2$$
$$K.E=\dfrac{1}{2}Iw^2$$ (1)
$$\therefore K-1=\dfrac{1}{2}\times Iw^2\implies I\propto\dfrac{1}{K.E}$$
$${ K }_{ 2 }=\frac { { MR }^{ 2 } }{ 2 } { w }^{ 2 }$$
Now $${ { K }_{ 1 }/K }_{ 2 }=\dfrac { m }{ m } \left( \dfrac { r }{ R } \right) ^{ 2 }$$ and $$K_1+K_2=U$$ (ii)
$$\implies K_2\times\dfrac{m}{M}(r/R)^2+K_2=U$$
$$\implies K_2=\dfrac{UMR^2}{mr^2+MR^2}$$
Similarly $${ K }_{ 1 }=\dfrac { UM{ R }^{ 2 } }{ { mr }^{ 2 }+{ MR }^{ 2 } } $$
And using (ii)
$$U=U=\dfrac { 1 }{ 2 } \frac { \pi { r }^{ 2 }{ \mu }_{ 2 }I }{ 2R } =\dfrac { { \mu }_{ 2 }{ I }^{ 2 }\pi { r }^{ 2 } }{ 4R } =U$$
Question 9
1 / -0

The coercivity of a magnet is $$3\times 10^{3}Am^{-1}$$. What current should be passed through a solenoid of length $$10\ cm$$ and number of turns $$50$$ that the magnet is demagnetized when inserted in it?

A
$$3A$$

B
$$6A$$

C
$$9A$$

D
$$12A$$

Solution

Length of solenoid   $$L = 10 \ cm = 0.1 \ m$$
Number of turns $$N = 50$$
So, number of turns per unit length   $$n = \dfrac{N}{L} = \dfrac{50}{0.1} = 500$$
Current to be passed   $$i = \dfrac{H}{n}$$
$$\implies \ i = \dfrac{3\times 10^3}{500} = 6 \ A$$
Question 10
1 / -0

A helium nucleus makes a full rotation of radius 0.8 metre in two seconds. The value of the magnetic field B at the centre of the circle will be:

A
$$\dfrac { { 10 }^{ -19 } }{ { \mu }_{ 0 } }$$

B
$${ 10 }^{ -19 }\ { \mu }_{ 0 }$$

C
$$2\ \times \ { 10 }^{ -10 }\ { \mu }_{ 0 }$$

D
$$\dfrac { 2\ \times \ { 10 }^{ -10 } }{ { \mu }_{ 0 } }$$

Solution

Given that Helium Nucleus makes a full rotation of a circle of radius 0.8m in 2 seconds.
So $$T = 2$$ seconds
$$r = 0.8$$ meter
Charge on the nucleus,
$$q = 2e = 2\times 1.6\times 10^{-19}C$$
$$= 3.2\times 10^{-19}C$$
Current produced by Helium nucleus,
$$I = \dfrac{q}{t} = \dfrac{3.2\times 10^{-31}C}{2} = 1.6\times 10^{-19}A$$
The magnetic field at the center of circle is given by
$$B = \dfrac{\mu_0}{4\pi} . \dfrac{2I}{r} = \dfrac{\mu_0I}{2r}$$
$$\Rightarrow B = \dfrac{\mu_0\times 1.6\times 10^{-19}}{2\times 0.8}$$
$$\Rightarrow B = \mu_0 \times 10^{-19}$$