154295
A domain in a ferromagnetic substance is in the form of a cube of length $1 \mu \mathrm{m}$. If it contains $8 \times$ $10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24} \mathrm{Am}^{2}$, then the magnetisation of the domain is
1 $7.2 \times 10^{5} \mathrm{Am}^{-1}$
2 $7.2 \times 10^{3} \mathrm{Am}^{-1}$
3 $7.2 \times 10^{9} \mathrm{Am}^{-1}$
4 $7.2 \times 10^{12} \mathrm{Am}^{-1}$
Explanation:
A Cube of side $=1 \mu \mathrm{m}=10^{-6} \mathrm{~m}$ Volume of cube $($ domain $)=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$ Number of atoms $(\mathrm{N})=8 \times 10^{10}$ Dipole moment (M) $=9 \times 10^{-24} \mathrm{Am}^{2}$ The maximum possible dipole moment $\mathrm{M}_{\max }$ is obtained for case when all atomic moment at perfectly aligned. Hence, $\mathrm{M}_{\max }=\mathrm{NM}=8 \times 10^{10} \times 9 \times 10^{-24}$ $=72 \times 10^{-14} \mathrm{Am}^{2}$ $\therefore \quad$ Magnetization $=\frac{\mathrm{M}_{\max }}{\text { Volume }}$ $=\frac{72 \times 10^{-14}}{10^{-18}}$ $=72 \times 10^{4}$ $=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$
MHT-CET 2020
Magnetism and Matter
154298
A bar magnet of magnetic moment $5 \mathrm{Am}^{2}$ is placed in a uniform magnetic induction $3 \times 10^{-5}$ T. If each pole of a magnet experiences a force of $2.5 \times 10^{-4} \mathrm{~N}$ then the magnetic length of the magnet is
1 $0.2 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.6 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Magnetic moment (M) $=5 \mathrm{Am}^{2}$ Magnetic induction $(\mathrm{B})=3 \times 10^{-5} \mathrm{~T}$ $\text { Force }(\mathrm{F})=2.5 \times 10^{-4} \mathrm{~N}$ $\because \mathrm{F}=\mathrm{mB}$ Where, $\mathrm{m}$ is pole strength. $\mathrm{m}=\frac{\mathrm{F}}{\mathrm{B}}=\frac{2.5 \times 10^{-4}}{3 \times 10^{-5}}=\frac{25}{3} \mathrm{Am}$ Magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{L}=\frac{\mathrm{M}}{\mathrm{m}}=\frac{5 \times 3}{25}=0.6 \mathrm{~m}$
MHT-CET 2020
Magnetism and Matter
154300
The relation between total magnetic field (B), magnetic intensity $(\mathrm{H})$, permeability of free space $\left(\mu_{0}\right)$ and susceptibility $(\chi)$ is
1 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1-\chi)$
2 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
3 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1+\chi)$
4 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1-\chi)$
Explanation:
B Magnetic susceptibility $(\chi)=\frac{\mathrm{I}}{\mathrm{H}}$ Magnetic permeability $(\mu)=\frac{B}{H}$ Let us calculate the net field inside a magnetic material, $\mathrm{B}_{\text {net }}=\mathrm{B}_{0}+\mathrm{B}_{\mathrm{i}}$ $\mathrm{B}=\mu_{0} \mathrm{H}+\mu_{0} \mathrm{I}$ $\mathrm{B}=\mu_{0}(\mathrm{H}+\mathrm{I})$ $\mathrm{B}=\mu_{\mathrm{o}}(\mathrm{H}+\chi \mathrm{H})$ $\{\because \mathrm{I}=\chi \mathrm{H}\}$ $\mathrm{B}=\mu_{0} \mathrm{H}(1+\chi)$ $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
MHT-CET 2020
Magnetism and Matter
154302
The susceptibility of tungsten is $6.8 \times 10^{-5}$ at temperature $300 \mathrm{~K}$. The susceptibility at temperature $400 \mathrm{~K}$ is
154295
A domain in a ferromagnetic substance is in the form of a cube of length $1 \mu \mathrm{m}$. If it contains $8 \times$ $10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24} \mathrm{Am}^{2}$, then the magnetisation of the domain is
1 $7.2 \times 10^{5} \mathrm{Am}^{-1}$
2 $7.2 \times 10^{3} \mathrm{Am}^{-1}$
3 $7.2 \times 10^{9} \mathrm{Am}^{-1}$
4 $7.2 \times 10^{12} \mathrm{Am}^{-1}$
Explanation:
A Cube of side $=1 \mu \mathrm{m}=10^{-6} \mathrm{~m}$ Volume of cube $($ domain $)=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$ Number of atoms $(\mathrm{N})=8 \times 10^{10}$ Dipole moment (M) $=9 \times 10^{-24} \mathrm{Am}^{2}$ The maximum possible dipole moment $\mathrm{M}_{\max }$ is obtained for case when all atomic moment at perfectly aligned. Hence, $\mathrm{M}_{\max }=\mathrm{NM}=8 \times 10^{10} \times 9 \times 10^{-24}$ $=72 \times 10^{-14} \mathrm{Am}^{2}$ $\therefore \quad$ Magnetization $=\frac{\mathrm{M}_{\max }}{\text { Volume }}$ $=\frac{72 \times 10^{-14}}{10^{-18}}$ $=72 \times 10^{4}$ $=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$
MHT-CET 2020
Magnetism and Matter
154298
A bar magnet of magnetic moment $5 \mathrm{Am}^{2}$ is placed in a uniform magnetic induction $3 \times 10^{-5}$ T. If each pole of a magnet experiences a force of $2.5 \times 10^{-4} \mathrm{~N}$ then the magnetic length of the magnet is
1 $0.2 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.6 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Magnetic moment (M) $=5 \mathrm{Am}^{2}$ Magnetic induction $(\mathrm{B})=3 \times 10^{-5} \mathrm{~T}$ $\text { Force }(\mathrm{F})=2.5 \times 10^{-4} \mathrm{~N}$ $\because \mathrm{F}=\mathrm{mB}$ Where, $\mathrm{m}$ is pole strength. $\mathrm{m}=\frac{\mathrm{F}}{\mathrm{B}}=\frac{2.5 \times 10^{-4}}{3 \times 10^{-5}}=\frac{25}{3} \mathrm{Am}$ Magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{L}=\frac{\mathrm{M}}{\mathrm{m}}=\frac{5 \times 3}{25}=0.6 \mathrm{~m}$
MHT-CET 2020
Magnetism and Matter
154300
The relation between total magnetic field (B), magnetic intensity $(\mathrm{H})$, permeability of free space $\left(\mu_{0}\right)$ and susceptibility $(\chi)$ is
1 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1-\chi)$
2 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
3 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1+\chi)$
4 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1-\chi)$
Explanation:
B Magnetic susceptibility $(\chi)=\frac{\mathrm{I}}{\mathrm{H}}$ Magnetic permeability $(\mu)=\frac{B}{H}$ Let us calculate the net field inside a magnetic material, $\mathrm{B}_{\text {net }}=\mathrm{B}_{0}+\mathrm{B}_{\mathrm{i}}$ $\mathrm{B}=\mu_{0} \mathrm{H}+\mu_{0} \mathrm{I}$ $\mathrm{B}=\mu_{0}(\mathrm{H}+\mathrm{I})$ $\mathrm{B}=\mu_{\mathrm{o}}(\mathrm{H}+\chi \mathrm{H})$ $\{\because \mathrm{I}=\chi \mathrm{H}\}$ $\mathrm{B}=\mu_{0} \mathrm{H}(1+\chi)$ $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
MHT-CET 2020
Magnetism and Matter
154302
The susceptibility of tungsten is $6.8 \times 10^{-5}$ at temperature $300 \mathrm{~K}$. The susceptibility at temperature $400 \mathrm{~K}$ is
154295
A domain in a ferromagnetic substance is in the form of a cube of length $1 \mu \mathrm{m}$. If it contains $8 \times$ $10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24} \mathrm{Am}^{2}$, then the magnetisation of the domain is
1 $7.2 \times 10^{5} \mathrm{Am}^{-1}$
2 $7.2 \times 10^{3} \mathrm{Am}^{-1}$
3 $7.2 \times 10^{9} \mathrm{Am}^{-1}$
4 $7.2 \times 10^{12} \mathrm{Am}^{-1}$
Explanation:
A Cube of side $=1 \mu \mathrm{m}=10^{-6} \mathrm{~m}$ Volume of cube $($ domain $)=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$ Number of atoms $(\mathrm{N})=8 \times 10^{10}$ Dipole moment (M) $=9 \times 10^{-24} \mathrm{Am}^{2}$ The maximum possible dipole moment $\mathrm{M}_{\max }$ is obtained for case when all atomic moment at perfectly aligned. Hence, $\mathrm{M}_{\max }=\mathrm{NM}=8 \times 10^{10} \times 9 \times 10^{-24}$ $=72 \times 10^{-14} \mathrm{Am}^{2}$ $\therefore \quad$ Magnetization $=\frac{\mathrm{M}_{\max }}{\text { Volume }}$ $=\frac{72 \times 10^{-14}}{10^{-18}}$ $=72 \times 10^{4}$ $=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$
MHT-CET 2020
Magnetism and Matter
154298
A bar magnet of magnetic moment $5 \mathrm{Am}^{2}$ is placed in a uniform magnetic induction $3 \times 10^{-5}$ T. If each pole of a magnet experiences a force of $2.5 \times 10^{-4} \mathrm{~N}$ then the magnetic length of the magnet is
1 $0.2 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.6 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Magnetic moment (M) $=5 \mathrm{Am}^{2}$ Magnetic induction $(\mathrm{B})=3 \times 10^{-5} \mathrm{~T}$ $\text { Force }(\mathrm{F})=2.5 \times 10^{-4} \mathrm{~N}$ $\because \mathrm{F}=\mathrm{mB}$ Where, $\mathrm{m}$ is pole strength. $\mathrm{m}=\frac{\mathrm{F}}{\mathrm{B}}=\frac{2.5 \times 10^{-4}}{3 \times 10^{-5}}=\frac{25}{3} \mathrm{Am}$ Magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{L}=\frac{\mathrm{M}}{\mathrm{m}}=\frac{5 \times 3}{25}=0.6 \mathrm{~m}$
MHT-CET 2020
Magnetism and Matter
154300
The relation between total magnetic field (B), magnetic intensity $(\mathrm{H})$, permeability of free space $\left(\mu_{0}\right)$ and susceptibility $(\chi)$ is
1 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1-\chi)$
2 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
3 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1+\chi)$
4 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1-\chi)$
Explanation:
B Magnetic susceptibility $(\chi)=\frac{\mathrm{I}}{\mathrm{H}}$ Magnetic permeability $(\mu)=\frac{B}{H}$ Let us calculate the net field inside a magnetic material, $\mathrm{B}_{\text {net }}=\mathrm{B}_{0}+\mathrm{B}_{\mathrm{i}}$ $\mathrm{B}=\mu_{0} \mathrm{H}+\mu_{0} \mathrm{I}$ $\mathrm{B}=\mu_{0}(\mathrm{H}+\mathrm{I})$ $\mathrm{B}=\mu_{\mathrm{o}}(\mathrm{H}+\chi \mathrm{H})$ $\{\because \mathrm{I}=\chi \mathrm{H}\}$ $\mathrm{B}=\mu_{0} \mathrm{H}(1+\chi)$ $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
MHT-CET 2020
Magnetism and Matter
154302
The susceptibility of tungsten is $6.8 \times 10^{-5}$ at temperature $300 \mathrm{~K}$. The susceptibility at temperature $400 \mathrm{~K}$ is
154295
A domain in a ferromagnetic substance is in the form of a cube of length $1 \mu \mathrm{m}$. If it contains $8 \times$ $10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24} \mathrm{Am}^{2}$, then the magnetisation of the domain is
1 $7.2 \times 10^{5} \mathrm{Am}^{-1}$
2 $7.2 \times 10^{3} \mathrm{Am}^{-1}$
3 $7.2 \times 10^{9} \mathrm{Am}^{-1}$
4 $7.2 \times 10^{12} \mathrm{Am}^{-1}$
Explanation:
A Cube of side $=1 \mu \mathrm{m}=10^{-6} \mathrm{~m}$ Volume of cube $($ domain $)=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$ Number of atoms $(\mathrm{N})=8 \times 10^{10}$ Dipole moment (M) $=9 \times 10^{-24} \mathrm{Am}^{2}$ The maximum possible dipole moment $\mathrm{M}_{\max }$ is obtained for case when all atomic moment at perfectly aligned. Hence, $\mathrm{M}_{\max }=\mathrm{NM}=8 \times 10^{10} \times 9 \times 10^{-24}$ $=72 \times 10^{-14} \mathrm{Am}^{2}$ $\therefore \quad$ Magnetization $=\frac{\mathrm{M}_{\max }}{\text { Volume }}$ $=\frac{72 \times 10^{-14}}{10^{-18}}$ $=72 \times 10^{4}$ $=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$
MHT-CET 2020
Magnetism and Matter
154298
A bar magnet of magnetic moment $5 \mathrm{Am}^{2}$ is placed in a uniform magnetic induction $3 \times 10^{-5}$ T. If each pole of a magnet experiences a force of $2.5 \times 10^{-4} \mathrm{~N}$ then the magnetic length of the magnet is
1 $0.2 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.6 \mathrm{~m}$
4 $0.4 \mathrm{~m}$
Explanation:
C Magnetic moment (M) $=5 \mathrm{Am}^{2}$ Magnetic induction $(\mathrm{B})=3 \times 10^{-5} \mathrm{~T}$ $\text { Force }(\mathrm{F})=2.5 \times 10^{-4} \mathrm{~N}$ $\because \mathrm{F}=\mathrm{mB}$ Where, $\mathrm{m}$ is pole strength. $\mathrm{m}=\frac{\mathrm{F}}{\mathrm{B}}=\frac{2.5 \times 10^{-4}}{3 \times 10^{-5}}=\frac{25}{3} \mathrm{Am}$ Magnetic moment $(\mathrm{M})=\mathrm{mL}$ $\mathrm{L}=\frac{\mathrm{M}}{\mathrm{m}}=\frac{5 \times 3}{25}=0.6 \mathrm{~m}$
MHT-CET 2020
Magnetism and Matter
154300
The relation between total magnetic field (B), magnetic intensity $(\mathrm{H})$, permeability of free space $\left(\mu_{0}\right)$ and susceptibility $(\chi)$ is
1 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1-\chi)$
2 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
3 $\frac{\mathrm{H}}{\mathrm{B}}=\mu_{0}(1+\chi)$
4 $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1-\chi)$
Explanation:
B Magnetic susceptibility $(\chi)=\frac{\mathrm{I}}{\mathrm{H}}$ Magnetic permeability $(\mu)=\frac{B}{H}$ Let us calculate the net field inside a magnetic material, $\mathrm{B}_{\text {net }}=\mathrm{B}_{0}+\mathrm{B}_{\mathrm{i}}$ $\mathrm{B}=\mu_{0} \mathrm{H}+\mu_{0} \mathrm{I}$ $\mathrm{B}=\mu_{0}(\mathrm{H}+\mathrm{I})$ $\mathrm{B}=\mu_{\mathrm{o}}(\mathrm{H}+\chi \mathrm{H})$ $\{\because \mathrm{I}=\chi \mathrm{H}\}$ $\mathrm{B}=\mu_{0} \mathrm{H}(1+\chi)$ $\frac{\mathrm{B}}{\mathrm{H}}=\mu_{0}(1+\chi)$
MHT-CET 2020
Magnetism and Matter
154302
The susceptibility of tungsten is $6.8 \times 10^{-5}$ at temperature $300 \mathrm{~K}$. The susceptibility at temperature $400 \mathrm{~K}$ is
