154913
The mass of a ferromagnetic material is $100 \mathrm{~g}$ and its magnetic moment is $10 \mathrm{~A} \mathrm{~m}^{2}$. If the density of the material is $10 \mathrm{~g} / \mathrm{cc}$, the intensity of magnetization is given by
1 $10^{8} \mathrm{~A} / \mathrm{m}$
2 $10^{6} \mathrm{~A} / \mathrm{m}$
3 $10^{4} \mathrm{~A} / \mathrm{m}$
4 $10^{2} \mathrm{~A} / \mathrm{m}$
Explanation:
B Given, Weight of material $(\mathrm{W})=100 \mathrm{~g}$ Magnetic moment $(\mathrm{M})=10 \mathrm{Am}^{2}$ Density of material $(\rho)=10 \mathrm{~g} / \mathrm{cc}=10 \times 10^{-6} \mathrm{~g} / \mathrm{m}^{3}$ We know, magnetization $\mathrm{B}=\frac{\text { Magnetic moment }}{\text { volume }}=\frac{\mathrm{M}}{\mathrm{V}}$ Desnity $=\frac{\text { Mass }}{\text { Volume }}$ Volume $=\frac{100}{10 \times 10^{6}}=10^{-5} \mathrm{~m}^{3}$ Putting the value of volume in equation (i), $\therefore \mathrm{B}=\frac{10}{10^{-5}}=10^{6} \mathrm{~A} / \mathrm{m}$ $\mathrm{B}=10^{6} \mathrm{~A} / \mathrm{m}$
SRMJEEE - 2011
Electro Magnetic Induction
154914
The energy stored in a coil of self inductance 40 mH carrying a steady current of $2 \mathrm{~A}$ is
1 $0.08 \mathrm{~J}$
2 $0.8 \mathrm{~J}$
3 $80 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
A Given, Self Inductance $(\mathrm{L})=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Energy store in a inductor $=\frac{1}{2} \mathrm{LI}^{2}=\frac{1}{2} 40 \times 10^{-3} \times(2)^{2}$ $=\frac{40 \times 10^{-3} \times 4}{2}=80 \times 10^{-3}$ $\mathrm{U} =0.08 \mathrm{~J}$
SRMJEEE - 2012
Electro Magnetic Induction
154915
If the inductance of a coil is 1 Henry, its effective resistance in a DC circuit will be
1 $\infty$
2 zero
3 $1 \Omega$
4 $2 \Omega$
Explanation:
B Given that, Inductance of coil $(\mathrm{L})=1 \mathrm{H}$ We know, effective inductive resistance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi f \mathrm{~L}$ $(\because \omega=2 \pi f)$ For DC,$f=0$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 0 \times \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=0$
154913
The mass of a ferromagnetic material is $100 \mathrm{~g}$ and its magnetic moment is $10 \mathrm{~A} \mathrm{~m}^{2}$. If the density of the material is $10 \mathrm{~g} / \mathrm{cc}$, the intensity of magnetization is given by
1 $10^{8} \mathrm{~A} / \mathrm{m}$
2 $10^{6} \mathrm{~A} / \mathrm{m}$
3 $10^{4} \mathrm{~A} / \mathrm{m}$
4 $10^{2} \mathrm{~A} / \mathrm{m}$
Explanation:
B Given, Weight of material $(\mathrm{W})=100 \mathrm{~g}$ Magnetic moment $(\mathrm{M})=10 \mathrm{Am}^{2}$ Density of material $(\rho)=10 \mathrm{~g} / \mathrm{cc}=10 \times 10^{-6} \mathrm{~g} / \mathrm{m}^{3}$ We know, magnetization $\mathrm{B}=\frac{\text { Magnetic moment }}{\text { volume }}=\frac{\mathrm{M}}{\mathrm{V}}$ Desnity $=\frac{\text { Mass }}{\text { Volume }}$ Volume $=\frac{100}{10 \times 10^{6}}=10^{-5} \mathrm{~m}^{3}$ Putting the value of volume in equation (i), $\therefore \mathrm{B}=\frac{10}{10^{-5}}=10^{6} \mathrm{~A} / \mathrm{m}$ $\mathrm{B}=10^{6} \mathrm{~A} / \mathrm{m}$
SRMJEEE - 2011
Electro Magnetic Induction
154914
The energy stored in a coil of self inductance 40 mH carrying a steady current of $2 \mathrm{~A}$ is
1 $0.08 \mathrm{~J}$
2 $0.8 \mathrm{~J}$
3 $80 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
A Given, Self Inductance $(\mathrm{L})=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Energy store in a inductor $=\frac{1}{2} \mathrm{LI}^{2}=\frac{1}{2} 40 \times 10^{-3} \times(2)^{2}$ $=\frac{40 \times 10^{-3} \times 4}{2}=80 \times 10^{-3}$ $\mathrm{U} =0.08 \mathrm{~J}$
SRMJEEE - 2012
Electro Magnetic Induction
154915
If the inductance of a coil is 1 Henry, its effective resistance in a DC circuit will be
1 $\infty$
2 zero
3 $1 \Omega$
4 $2 \Omega$
Explanation:
B Given that, Inductance of coil $(\mathrm{L})=1 \mathrm{H}$ We know, effective inductive resistance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi f \mathrm{~L}$ $(\because \omega=2 \pi f)$ For DC,$f=0$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 0 \times \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=0$
154913
The mass of a ferromagnetic material is $100 \mathrm{~g}$ and its magnetic moment is $10 \mathrm{~A} \mathrm{~m}^{2}$. If the density of the material is $10 \mathrm{~g} / \mathrm{cc}$, the intensity of magnetization is given by
1 $10^{8} \mathrm{~A} / \mathrm{m}$
2 $10^{6} \mathrm{~A} / \mathrm{m}$
3 $10^{4} \mathrm{~A} / \mathrm{m}$
4 $10^{2} \mathrm{~A} / \mathrm{m}$
Explanation:
B Given, Weight of material $(\mathrm{W})=100 \mathrm{~g}$ Magnetic moment $(\mathrm{M})=10 \mathrm{Am}^{2}$ Density of material $(\rho)=10 \mathrm{~g} / \mathrm{cc}=10 \times 10^{-6} \mathrm{~g} / \mathrm{m}^{3}$ We know, magnetization $\mathrm{B}=\frac{\text { Magnetic moment }}{\text { volume }}=\frac{\mathrm{M}}{\mathrm{V}}$ Desnity $=\frac{\text { Mass }}{\text { Volume }}$ Volume $=\frac{100}{10 \times 10^{6}}=10^{-5} \mathrm{~m}^{3}$ Putting the value of volume in equation (i), $\therefore \mathrm{B}=\frac{10}{10^{-5}}=10^{6} \mathrm{~A} / \mathrm{m}$ $\mathrm{B}=10^{6} \mathrm{~A} / \mathrm{m}$
SRMJEEE - 2011
Electro Magnetic Induction
154914
The energy stored in a coil of self inductance 40 mH carrying a steady current of $2 \mathrm{~A}$ is
1 $0.08 \mathrm{~J}$
2 $0.8 \mathrm{~J}$
3 $80 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
A Given, Self Inductance $(\mathrm{L})=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Energy store in a inductor $=\frac{1}{2} \mathrm{LI}^{2}=\frac{1}{2} 40 \times 10^{-3} \times(2)^{2}$ $=\frac{40 \times 10^{-3} \times 4}{2}=80 \times 10^{-3}$ $\mathrm{U} =0.08 \mathrm{~J}$
SRMJEEE - 2012
Electro Magnetic Induction
154915
If the inductance of a coil is 1 Henry, its effective resistance in a DC circuit will be
1 $\infty$
2 zero
3 $1 \Omega$
4 $2 \Omega$
Explanation:
B Given that, Inductance of coil $(\mathrm{L})=1 \mathrm{H}$ We know, effective inductive resistance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi f \mathrm{~L}$ $(\because \omega=2 \pi f)$ For DC,$f=0$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 0 \times \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=0$
154913
The mass of a ferromagnetic material is $100 \mathrm{~g}$ and its magnetic moment is $10 \mathrm{~A} \mathrm{~m}^{2}$. If the density of the material is $10 \mathrm{~g} / \mathrm{cc}$, the intensity of magnetization is given by
1 $10^{8} \mathrm{~A} / \mathrm{m}$
2 $10^{6} \mathrm{~A} / \mathrm{m}$
3 $10^{4} \mathrm{~A} / \mathrm{m}$
4 $10^{2} \mathrm{~A} / \mathrm{m}$
Explanation:
B Given, Weight of material $(\mathrm{W})=100 \mathrm{~g}$ Magnetic moment $(\mathrm{M})=10 \mathrm{Am}^{2}$ Density of material $(\rho)=10 \mathrm{~g} / \mathrm{cc}=10 \times 10^{-6} \mathrm{~g} / \mathrm{m}^{3}$ We know, magnetization $\mathrm{B}=\frac{\text { Magnetic moment }}{\text { volume }}=\frac{\mathrm{M}}{\mathrm{V}}$ Desnity $=\frac{\text { Mass }}{\text { Volume }}$ Volume $=\frac{100}{10 \times 10^{6}}=10^{-5} \mathrm{~m}^{3}$ Putting the value of volume in equation (i), $\therefore \mathrm{B}=\frac{10}{10^{-5}}=10^{6} \mathrm{~A} / \mathrm{m}$ $\mathrm{B}=10^{6} \mathrm{~A} / \mathrm{m}$
SRMJEEE - 2011
Electro Magnetic Induction
154914
The energy stored in a coil of self inductance 40 mH carrying a steady current of $2 \mathrm{~A}$ is
1 $0.08 \mathrm{~J}$
2 $0.8 \mathrm{~J}$
3 $80 \mathrm{~J}$
4 $8 \mathrm{~J}$
Explanation:
A Given, Self Inductance $(\mathrm{L})=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Current $(\mathrm{I})=2 \mathrm{~A}$ Energy store in a inductor $=\frac{1}{2} \mathrm{LI}^{2}=\frac{1}{2} 40 \times 10^{-3} \times(2)^{2}$ $=\frac{40 \times 10^{-3} \times 4}{2}=80 \times 10^{-3}$ $\mathrm{U} =0.08 \mathrm{~J}$
SRMJEEE - 2012
Electro Magnetic Induction
154915
If the inductance of a coil is 1 Henry, its effective resistance in a DC circuit will be
1 $\infty$
2 zero
3 $1 \Omega$
4 $2 \Omega$
Explanation:
B Given that, Inductance of coil $(\mathrm{L})=1 \mathrm{H}$ We know, effective inductive resistance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=2 \pi f \mathrm{~L}$ $(\because \omega=2 \pi f)$ For DC,$f=0$ $\mathrm{X}_{\mathrm{L}}=2 \pi \times 0 \times \mathrm{L}$ $\mathrm{X}_{\mathrm{L}}=0$
