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Magnetism and Matter

154321 The magnetization of bar magnet of length 5 cm, cross sectional area $2 {cm}^{2}$ net magnetic moment $1 {Am}^{2}$ is

1

2 \times 10^{5} A / m

2

3 \times 10^{5} A / m

3

1 \times 10^{5} A / m

4

4 \times 10^{5} A / m

Explanation:

C Given,
Length $(L) = 5 cm = 5 \times 10^{- 2} m$ ,
Area $(A) = 2 {cm}^{2} = 2 \times 10^{- 4} m^{2}$
Magnetic moment $(M) = 1 {Am}^{- 2}$
The magnetisation of the bar magnet is
$I = \frac{M}{V}$
$I = \frac{M}{A \times L} (∴ V = A \times L)$
Substituting given values, we get
$I = \frac{1}{2 \times 10^{- 4} \times 5 \times 10^{- 2}} = \frac{1}{10^{- 5}}$
$= 1 \times 10^{5} Am$

MHT-CET 2019

Magnetism and Matter

154322 A magnetizing field of $5000 A / m$ produces a magnetic flux of $4 \times 10^{- 5}$ weber in an iron rod of cross sectional area $0.4 {cm}^{2}$ . The permeability of the $rod$ in $Wb / Am$ is

1

2 \times 10^{- 4}

2

4 \times 10^{- 6}

3

3 \times 10^{- 5}

4

1 \times 10^{- 3}

Explanation:

A Given,
$H = 5000 A / m$
$ϕ = 4 \times 10^{- 5} wb$
$A = 0.4 {cm}^{2} = 4 \times 10^{- 5} m^{2}$
$μ = \frac{B}{H} = \frac{ϕ}{AH} (∵ ϕ = BA)$
$= \frac{4 \times 10^{- 5}}{4 \times 10^{- 5} \times 5000} = 2 \times 10^{- 4} Wb / A - m$

MHT-CET 2019

Magnetism and Matter

154326 The magnetic susceptibility of a paramagnetic substance at $- 173^{\circ} C$ is $1.5 \times 10^{- 2}$ . To have the susceptibility $0.5 \times 10^{- 2}$ , the change in temperature in $^{\circ} C$ is

1 100

2 180

3 200

4 220

Explanation:

C Given,
$T_{1} = - 173^{\circ} C = (- 173 + 273) = 100 K$
$χ_{m_{1}} = 1.5 \times 10^{- 2}$
$χ_{m_{2}} = 0.5 \times 10^{- 2}$
According to curie's law,
$χ_{m} \propto \frac{1}{T}$
$\frac{χ_{m_{1}}}{χ_{m_{2}}} = \frac{T_{2}}{T_{1}}$
$T_{2} = \frac{χ_{m_{1}} T_{1}}{χ_{m_{2}}}$
$= \frac{1.5 \times 10^{- 2} \times 100}{0.5 \times 10^{- 2}} = 300 K$
$∴ T_{2} = (300 - 273) = 27^{\circ} C$
Hence, change in temperature,
$Δ T = T_{2} - T_{1}$
$= 27 - (- 173) = 200^{\circ} C$

AP EAMCET (21.04.2019) Shift-II

Magnetism and Matter

154328 The B-H curves for two ferromagnetic materials are shows in figure.

1 (1) soft iron and (2) steel

2 (1) steel and (2) soft iron

3 (1) diamagnetic and (2) paramagnetic

4 (1) paramagnetic and (2) ferromagnetic
[SRM JEE-2018]

Explanation:

A The loop $- I$ is narrower than loop -II, retentively is more in case of steel.
Hence, loop-I is for soft iron and loop - II is for steel.

Magnetism and Matter

154321 The magnetization of bar magnet of length 5 cm, cross sectional area $2 {cm}^{2}$ net magnetic moment $1 {Am}^{2}$ is

1

2 \times 10^{5} A / m

2

3 \times 10^{5} A / m

3

1 \times 10^{5} A / m

4

4 \times 10^{5} A / m

Explanation:

C Given,
Length $(L) = 5 cm = 5 \times 10^{- 2} m$ ,
Area $(A) = 2 {cm}^{2} = 2 \times 10^{- 4} m^{2}$
Magnetic moment $(M) = 1 {Am}^{- 2}$
The magnetisation of the bar magnet is
$I = \frac{M}{V}$
$I = \frac{M}{A \times L} (∴ V = A \times L)$
Substituting given values, we get
$I = \frac{1}{2 \times 10^{- 4} \times 5 \times 10^{- 2}} = \frac{1}{10^{- 5}}$
$= 1 \times 10^{5} Am$

MHT-CET 2019

Magnetism and Matter

154322 A magnetizing field of $5000 A / m$ produces a magnetic flux of $4 \times 10^{- 5}$ weber in an iron rod of cross sectional area $0.4 {cm}^{2}$ . The permeability of the $rod$ in $Wb / Am$ is

1

2 \times 10^{- 4}

2

4 \times 10^{- 6}

3

3 \times 10^{- 5}

4

1 \times 10^{- 3}

Explanation:

A Given,
$H = 5000 A / m$
$ϕ = 4 \times 10^{- 5} wb$
$A = 0.4 {cm}^{2} = 4 \times 10^{- 5} m^{2}$
$μ = \frac{B}{H} = \frac{ϕ}{AH} (∵ ϕ = BA)$
$= \frac{4 \times 10^{- 5}}{4 \times 10^{- 5} \times 5000} = 2 \times 10^{- 4} Wb / A - m$

MHT-CET 2019

Magnetism and Matter

154326 The magnetic susceptibility of a paramagnetic substance at $- 173^{\circ} C$ is $1.5 \times 10^{- 2}$ . To have the susceptibility $0.5 \times 10^{- 2}$ , the change in temperature in $^{\circ} C$ is

1 100

2 180

3 200

4 220

Explanation:

C Given,
$T_{1} = - 173^{\circ} C = (- 173 + 273) = 100 K$
$χ_{m_{1}} = 1.5 \times 10^{- 2}$
$χ_{m_{2}} = 0.5 \times 10^{- 2}$
According to curie's law,
$χ_{m} \propto \frac{1}{T}$
$\frac{χ_{m_{1}}}{χ_{m_{2}}} = \frac{T_{2}}{T_{1}}$
$T_{2} = \frac{χ_{m_{1}} T_{1}}{χ_{m_{2}}}$
$= \frac{1.5 \times 10^{- 2} \times 100}{0.5 \times 10^{- 2}} = 300 K$
$∴ T_{2} = (300 - 273) = 27^{\circ} C$
Hence, change in temperature,
$Δ T = T_{2} - T_{1}$
$= 27 - (- 173) = 200^{\circ} C$

AP EAMCET (21.04.2019) Shift-II

Magnetism and Matter

154328 The B-H curves for two ferromagnetic materials are shows in figure.

1 (1) soft iron and (2) steel

2 (1) steel and (2) soft iron

3 (1) diamagnetic and (2) paramagnetic

4 (1) paramagnetic and (2) ferromagnetic
[SRM JEE-2018]

Explanation:

A The loop $- I$ is narrower than loop -II, retentively is more in case of steel.
Hence, loop-I is for soft iron and loop - II is for steel.

Magnetism and Matter

154321 The magnetization of bar magnet of length 5 cm, cross sectional area $2 {cm}^{2}$ net magnetic moment $1 {Am}^{2}$ is

1

2 \times 10^{5} A / m

2

3 \times 10^{5} A / m

3

1 \times 10^{5} A / m

4

4 \times 10^{5} A / m

Explanation:

C Given,
Length $(L) = 5 cm = 5 \times 10^{- 2} m$ ,
Area $(A) = 2 {cm}^{2} = 2 \times 10^{- 4} m^{2}$
Magnetic moment $(M) = 1 {Am}^{- 2}$
The magnetisation of the bar magnet is
$I = \frac{M}{V}$
$I = \frac{M}{A \times L} (∴ V = A \times L)$
Substituting given values, we get
$I = \frac{1}{2 \times 10^{- 4} \times 5 \times 10^{- 2}} = \frac{1}{10^{- 5}}$
$= 1 \times 10^{5} Am$

MHT-CET 2019

Magnetism and Matter

154322 A magnetizing field of $5000 A / m$ produces a magnetic flux of $4 \times 10^{- 5}$ weber in an iron rod of cross sectional area $0.4 {cm}^{2}$ . The permeability of the $rod$ in $Wb / Am$ is

1

2 \times 10^{- 4}

2

4 \times 10^{- 6}

3

3 \times 10^{- 5}

4

1 \times 10^{- 3}

Explanation:

A Given,
$H = 5000 A / m$
$ϕ = 4 \times 10^{- 5} wb$
$A = 0.4 {cm}^{2} = 4 \times 10^{- 5} m^{2}$
$μ = \frac{B}{H} = \frac{ϕ}{AH} (∵ ϕ = BA)$
$= \frac{4 \times 10^{- 5}}{4 \times 10^{- 5} \times 5000} = 2 \times 10^{- 4} Wb / A - m$

MHT-CET 2019

Magnetism and Matter

154326 The magnetic susceptibility of a paramagnetic substance at $- 173^{\circ} C$ is $1.5 \times 10^{- 2}$ . To have the susceptibility $0.5 \times 10^{- 2}$ , the change in temperature in $^{\circ} C$ is

1 100

2 180

3 200

4 220

Explanation:

C Given,
$T_{1} = - 173^{\circ} C = (- 173 + 273) = 100 K$
$χ_{m_{1}} = 1.5 \times 10^{- 2}$
$χ_{m_{2}} = 0.5 \times 10^{- 2}$
According to curie's law,
$χ_{m} \propto \frac{1}{T}$
$\frac{χ_{m_{1}}}{χ_{m_{2}}} = \frac{T_{2}}{T_{1}}$
$T_{2} = \frac{χ_{m_{1}} T_{1}}{χ_{m_{2}}}$
$= \frac{1.5 \times 10^{- 2} \times 100}{0.5 \times 10^{- 2}} = 300 K$
$∴ T_{2} = (300 - 273) = 27^{\circ} C$
Hence, change in temperature,
$Δ T = T_{2} - T_{1}$
$= 27 - (- 173) = 200^{\circ} C$

AP EAMCET (21.04.2019) Shift-II

Magnetism and Matter

154328 The B-H curves for two ferromagnetic materials are shows in figure.

1 (1) soft iron and (2) steel

2 (1) steel and (2) soft iron

3 (1) diamagnetic and (2) paramagnetic

4 (1) paramagnetic and (2) ferromagnetic
[SRM JEE-2018]

Explanation:

A The loop $- I$ is narrower than loop -II, retentively is more in case of steel.
Hence, loop-I is for soft iron and loop - II is for steel.

Magnetism and Matter

154321 The magnetization of bar magnet of length 5 cm, cross sectional area $2 {cm}^{2}$ net magnetic moment $1 {Am}^{2}$ is

1

2 \times 10^{5} A / m

2

3 \times 10^{5} A / m

3

1 \times 10^{5} A / m

4

4 \times 10^{5} A / m

Explanation:

C Given,
Length $(L) = 5 cm = 5 \times 10^{- 2} m$ ,
Area $(A) = 2 {cm}^{2} = 2 \times 10^{- 4} m^{2}$
Magnetic moment $(M) = 1 {Am}^{- 2}$
The magnetisation of the bar magnet is
$I = \frac{M}{V}$
$I = \frac{M}{A \times L} (∴ V = A \times L)$
Substituting given values, we get
$I = \frac{1}{2 \times 10^{- 4} \times 5 \times 10^{- 2}} = \frac{1}{10^{- 5}}$
$= 1 \times 10^{5} Am$

MHT-CET 2019

Magnetism and Matter

154322 A magnetizing field of $5000 A / m$ produces a magnetic flux of $4 \times 10^{- 5}$ weber in an iron rod of cross sectional area $0.4 {cm}^{2}$ . The permeability of the $rod$ in $Wb / Am$ is

1

2 \times 10^{- 4}

2

4 \times 10^{- 6}

3

3 \times 10^{- 5}

4

1 \times 10^{- 3}

Explanation:

A Given,
$H = 5000 A / m$
$ϕ = 4 \times 10^{- 5} wb$
$A = 0.4 {cm}^{2} = 4 \times 10^{- 5} m^{2}$
$μ = \frac{B}{H} = \frac{ϕ}{AH} (∵ ϕ = BA)$
$= \frac{4 \times 10^{- 5}}{4 \times 10^{- 5} \times 5000} = 2 \times 10^{- 4} Wb / A - m$

MHT-CET 2019

Magnetism and Matter

154326 The magnetic susceptibility of a paramagnetic substance at $- 173^{\circ} C$ is $1.5 \times 10^{- 2}$ . To have the susceptibility $0.5 \times 10^{- 2}$ , the change in temperature in $^{\circ} C$ is

1 100

2 180

3 200

4 220

Explanation:

C Given,
$T_{1} = - 173^{\circ} C = (- 173 + 273) = 100 K$
$χ_{m_{1}} = 1.5 \times 10^{- 2}$
$χ_{m_{2}} = 0.5 \times 10^{- 2}$
According to curie's law,
$χ_{m} \propto \frac{1}{T}$
$\frac{χ_{m_{1}}}{χ_{m_{2}}} = \frac{T_{2}}{T_{1}}$
$T_{2} = \frac{χ_{m_{1}} T_{1}}{χ_{m_{2}}}$
$= \frac{1.5 \times 10^{- 2} \times 100}{0.5 \times 10^{- 2}} = 300 K$
$∴ T_{2} = (300 - 273) = 27^{\circ} C$
Hence, change in temperature,
$Δ T = T_{2} - T_{1}$
$= 27 - (- 173) = 200^{\circ} C$

AP EAMCET (21.04.2019) Shift-II

Magnetism and Matter

154328 The B-H curves for two ferromagnetic materials are shows in figure.

1 (1) soft iron and (2) steel

2 (1) steel and (2) soft iron

3 (1) diamagnetic and (2) paramagnetic

4 (1) paramagnetic and (2) ferromagnetic
[SRM JEE-2018]

Explanation:

A The loop $- I$ is narrower than loop -II, retentively is more in case of steel.
Hence, loop-I is for soft iron and loop - II is for steel.

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