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Magnetism and Matter

154315 In an atom, electron of charge (-e) perform U.C.M. around a stationary positively charged nucleus, with period of revolution ' $T$ '. If ' $r$ ' is the radius of the orbit of the electron and ' $v$ ' is the orbital velocity, then the circulating current (I) is proportional to

1

e^{1} r^{- 1} v^{- 1}

2

e^{1} v^{1} r^{- 1}

3

v^{1} r^{1} e^{- 1}

4

e^{1} r^{1} v^{- 1}

Explanation:

B Given, charge on electron $= e^{-}$ , radius of orbit $= r$ , orbital velocity $= v$
$∴ v = \frac{d}{t}$
For circular orbit
$d = 2 π r$ (circumference of circle)
$v = \frac{2 π r}{t} \Rightarrow t = \frac{2 π r}{v}$
$∵ i = \frac{q}{t} = \frac{e}{t}$
$i = \frac{\frac{e}{2 π r}}{v} = \frac{e v}{2 π r}$
$i = \frac{e^{1} v^{1} r^{- 1}}{2 π}$
$i \propto e^{1} v^{1} r^{- 1}$

MHT-CET 2020

Magnetism and Matter

154317 Two identical thin bar magnets, each of length $l$ and pole strength $m$ , are placed at right angle to each other with north pole of one touching south pole of the other. The magnetic moment of the system is

1

0.5 lm

2

l m

3

2 lm

4

\sqrt{2} lm

Explanation:

D Given that,
Length of bar magnet $= L$
Pole strength $= m$
The angle between magnetic moments is, $θ = 90^{\circ}$
Let magnetic moment of bar $(M) = m l$
Resultant magnetic moment
$= \sqrt{M_{1}^{2} + M_{2}^{2} + 2 M_{1} M_{2} \cos θ} .$
$= \sqrt{{(M_{1})}^{2} + {(M_{2})}^{2}} (∵ M_{1} = M_{2} = M)$
$= \sqrt{M^{2} + M^{2}} = \sqrt{2} M = \sqrt{2} m l$

AP EAMCET (21.09.2020) Shift-I

Magnetism and Matter

154318 If relative permeability of iron is 5500 , then its susceptibility is

1

5500 \times 10^{7}

2

5500 \times 10^{- 7}

3 5501

4 5499

Explanation:

D Given that,
$μ_{r} = 5500$
We know that, relation between relative permeability and magnetic susceptibility is given by,
$μ_{r} = 1 + χ_{m}$
$χ_{m} = μ_{r} - 1$
$= 5500 - 1$
$= 5499$

TS- EAMCET-10.09.2020

Magnetism and Matter

154320 Coersivity of a magnet where the ferromagnet gets completely demagnetized is $3 \times 10^{3} {Am}^{- 1}$ . The minimum current required to be passes in a solenoid having 1000 turns per metre, so that the magnet gets completely damagnetized when placed inside the solenoid is :

1

30 mA

2

60 mA

3

3 A

4

6 A

Explanation:

C The intensity of the magnetic field at which the magnet is demagnetized is called coercivety of the magnet
$Given, H = 3 \times 10^{- 3} {Am}^{- 1}$
$n = 1000 turns / m$
For a solenoid,
$B = μ_{0} nI$
$μ_{0} H = μ_{0} nI$
Or $H = nI$
Putting these value, we get -
$3 \times 10^{3} = 1000 \times I$
$I = \frac{3 \times 10^{3}}{1000} = 3 A$

Karnataka CET-2019

Magnetism and Matter

154315 In an atom, electron of charge (-e) perform U.C.M. around a stationary positively charged nucleus, with period of revolution ' $T$ '. If ' $r$ ' is the radius of the orbit of the electron and ' $v$ ' is the orbital velocity, then the circulating current (I) is proportional to

1

e^{1} r^{- 1} v^{- 1}

2

e^{1} v^{1} r^{- 1}

3

v^{1} r^{1} e^{- 1}

4

e^{1} r^{1} v^{- 1}

Explanation:

B Given, charge on electron $= e^{-}$ , radius of orbit $= r$ , orbital velocity $= v$
$∴ v = \frac{d}{t}$
For circular orbit
$d = 2 π r$ (circumference of circle)
$v = \frac{2 π r}{t} \Rightarrow t = \frac{2 π r}{v}$
$∵ i = \frac{q}{t} = \frac{e}{t}$
$i = \frac{\frac{e}{2 π r}}{v} = \frac{e v}{2 π r}$
$i = \frac{e^{1} v^{1} r^{- 1}}{2 π}$
$i \propto e^{1} v^{1} r^{- 1}$

MHT-CET 2020

Magnetism and Matter

154317 Two identical thin bar magnets, each of length $l$ and pole strength $m$ , are placed at right angle to each other with north pole of one touching south pole of the other. The magnetic moment of the system is

1

0.5 lm

2

l m

3

2 lm

4

\sqrt{2} lm

Explanation:

D Given that,
Length of bar magnet $= L$
Pole strength $= m$
The angle between magnetic moments is, $θ = 90^{\circ}$
Let magnetic moment of bar $(M) = m l$
Resultant magnetic moment
$= \sqrt{M_{1}^{2} + M_{2}^{2} + 2 M_{1} M_{2} \cos θ} .$
$= \sqrt{{(M_{1})}^{2} + {(M_{2})}^{2}} (∵ M_{1} = M_{2} = M)$
$= \sqrt{M^{2} + M^{2}} = \sqrt{2} M = \sqrt{2} m l$

AP EAMCET (21.09.2020) Shift-I

Magnetism and Matter

154318 If relative permeability of iron is 5500 , then its susceptibility is

1

5500 \times 10^{7}

2

5500 \times 10^{- 7}

3 5501

4 5499

Explanation:

D Given that,
$μ_{r} = 5500$
We know that, relation between relative permeability and magnetic susceptibility is given by,
$μ_{r} = 1 + χ_{m}$
$χ_{m} = μ_{r} - 1$
$= 5500 - 1$
$= 5499$

TS- EAMCET-10.09.2020

Magnetism and Matter

154320 Coersivity of a magnet where the ferromagnet gets completely demagnetized is $3 \times 10^{3} {Am}^{- 1}$ . The minimum current required to be passes in a solenoid having 1000 turns per metre, so that the magnet gets completely damagnetized when placed inside the solenoid is :

1

30 mA

2

60 mA

3

3 A

4

6 A

Explanation:

C The intensity of the magnetic field at which the magnet is demagnetized is called coercivety of the magnet
$Given, H = 3 \times 10^{- 3} {Am}^{- 1}$
$n = 1000 turns / m$
For a solenoid,
$B = μ_{0} nI$
$μ_{0} H = μ_{0} nI$
Or $H = nI$
Putting these value, we get -
$3 \times 10^{3} = 1000 \times I$
$I = \frac{3 \times 10^{3}}{1000} = 3 A$

Karnataka CET-2019

Magnetism and Matter

154315 In an atom, electron of charge (-e) perform U.C.M. around a stationary positively charged nucleus, with period of revolution ' $T$ '. If ' $r$ ' is the radius of the orbit of the electron and ' $v$ ' is the orbital velocity, then the circulating current (I) is proportional to

1

e^{1} r^{- 1} v^{- 1}

2

e^{1} v^{1} r^{- 1}

3

v^{1} r^{1} e^{- 1}

4

e^{1} r^{1} v^{- 1}

Explanation:

B Given, charge on electron $= e^{-}$ , radius of orbit $= r$ , orbital velocity $= v$
$∴ v = \frac{d}{t}$
For circular orbit
$d = 2 π r$ (circumference of circle)
$v = \frac{2 π r}{t} \Rightarrow t = \frac{2 π r}{v}$
$∵ i = \frac{q}{t} = \frac{e}{t}$
$i = \frac{\frac{e}{2 π r}}{v} = \frac{e v}{2 π r}$
$i = \frac{e^{1} v^{1} r^{- 1}}{2 π}$
$i \propto e^{1} v^{1} r^{- 1}$

MHT-CET 2020

Magnetism and Matter

154317 Two identical thin bar magnets, each of length $l$ and pole strength $m$ , are placed at right angle to each other with north pole of one touching south pole of the other. The magnetic moment of the system is

1

0.5 lm

2

l m

3

2 lm

4

\sqrt{2} lm

Explanation:

D Given that,
Length of bar magnet $= L$
Pole strength $= m$
The angle between magnetic moments is, $θ = 90^{\circ}$
Let magnetic moment of bar $(M) = m l$
Resultant magnetic moment
$= \sqrt{M_{1}^{2} + M_{2}^{2} + 2 M_{1} M_{2} \cos θ} .$
$= \sqrt{{(M_{1})}^{2} + {(M_{2})}^{2}} (∵ M_{1} = M_{2} = M)$
$= \sqrt{M^{2} + M^{2}} = \sqrt{2} M = \sqrt{2} m l$

AP EAMCET (21.09.2020) Shift-I

Magnetism and Matter

154318 If relative permeability of iron is 5500 , then its susceptibility is

1

5500 \times 10^{7}

2

5500 \times 10^{- 7}

3 5501

4 5499

Explanation:

D Given that,
$μ_{r} = 5500$
We know that, relation between relative permeability and magnetic susceptibility is given by,
$μ_{r} = 1 + χ_{m}$
$χ_{m} = μ_{r} - 1$
$= 5500 - 1$
$= 5499$

TS- EAMCET-10.09.2020

Magnetism and Matter

154320 Coersivity of a magnet where the ferromagnet gets completely demagnetized is $3 \times 10^{3} {Am}^{- 1}$ . The minimum current required to be passes in a solenoid having 1000 turns per metre, so that the magnet gets completely damagnetized when placed inside the solenoid is :

1

30 mA

2

60 mA

3

3 A

4

6 A

Explanation:

C The intensity of the magnetic field at which the magnet is demagnetized is called coercivety of the magnet
$Given, H = 3 \times 10^{- 3} {Am}^{- 1}$
$n = 1000 turns / m$
For a solenoid,
$B = μ_{0} nI$
$μ_{0} H = μ_{0} nI$
Or $H = nI$
Putting these value, we get -
$3 \times 10^{3} = 1000 \times I$
$I = \frac{3 \times 10^{3}}{1000} = 3 A$

Karnataka CET-2019

Magnetism and Matter

154315 In an atom, electron of charge (-e) perform U.C.M. around a stationary positively charged nucleus, with period of revolution ' $T$ '. If ' $r$ ' is the radius of the orbit of the electron and ' $v$ ' is the orbital velocity, then the circulating current (I) is proportional to

1

e^{1} r^{- 1} v^{- 1}

2

e^{1} v^{1} r^{- 1}

3

v^{1} r^{1} e^{- 1}

4

e^{1} r^{1} v^{- 1}

Explanation:

B Given, charge on electron $= e^{-}$ , radius of orbit $= r$ , orbital velocity $= v$
$∴ v = \frac{d}{t}$
For circular orbit
$d = 2 π r$ (circumference of circle)
$v = \frac{2 π r}{t} \Rightarrow t = \frac{2 π r}{v}$
$∵ i = \frac{q}{t} = \frac{e}{t}$
$i = \frac{\frac{e}{2 π r}}{v} = \frac{e v}{2 π r}$
$i = \frac{e^{1} v^{1} r^{- 1}}{2 π}$
$i \propto e^{1} v^{1} r^{- 1}$

MHT-CET 2020

Magnetism and Matter

154317 Two identical thin bar magnets, each of length $l$ and pole strength $m$ , are placed at right angle to each other with north pole of one touching south pole of the other. The magnetic moment of the system is

1

0.5 lm

2

l m

3

2 lm

4

\sqrt{2} lm

Explanation:

D Given that,
Length of bar magnet $= L$
Pole strength $= m$
The angle between magnetic moments is, $θ = 90^{\circ}$
Let magnetic moment of bar $(M) = m l$
Resultant magnetic moment
$= \sqrt{M_{1}^{2} + M_{2}^{2} + 2 M_{1} M_{2} \cos θ} .$
$= \sqrt{{(M_{1})}^{2} + {(M_{2})}^{2}} (∵ M_{1} = M_{2} = M)$
$= \sqrt{M^{2} + M^{2}} = \sqrt{2} M = \sqrt{2} m l$

AP EAMCET (21.09.2020) Shift-I

Magnetism and Matter

154318 If relative permeability of iron is 5500 , then its susceptibility is

1

5500 \times 10^{7}

2

5500 \times 10^{- 7}

3 5501

4 5499

Explanation:

D Given that,
$μ_{r} = 5500$
We know that, relation between relative permeability and magnetic susceptibility is given by,
$μ_{r} = 1 + χ_{m}$
$χ_{m} = μ_{r} - 1$
$= 5500 - 1$
$= 5499$

TS- EAMCET-10.09.2020

Magnetism and Matter

154320 Coersivity of a magnet where the ferromagnet gets completely demagnetized is $3 \times 10^{3} {Am}^{- 1}$ . The minimum current required to be passes in a solenoid having 1000 turns per metre, so that the magnet gets completely damagnetized when placed inside the solenoid is :

1

30 mA

2

60 mA

3

3 A

4

6 A

Explanation:

C The intensity of the magnetic field at which the magnet is demagnetized is called coercivety of the magnet
$Given, H = 3 \times 10^{- 3} {Am}^{- 1}$
$n = 1000 turns / m$
For a solenoid,
$B = μ_{0} nI$
$μ_{0} H = μ_{0} nI$
Or $H = nI$
Putting these value, we get -
$3 \times 10^{3} = 1000 \times I$
$I = \frac{3 \times 10^{3}}{1000} = 3 A$

Karnataka CET-2019

Question Bank Series

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