154722
The magnetic energy stored in an inductor of inductance $4 \mu \mathrm{H}$ carrying a current of $2 \mathrm{~A}$ is:
1 $4 \mathrm{~mJ}$
2 $8 \mathrm{~mJ}$
3 $8 \mu \mathrm{J}$
4 $4 \mu \mathrm{J}$
Explanation:
C Given, Inductance $(\mathrm{L})=4 \mu \mathrm{H}$ Current (i) $=2 \mathrm{~A}$ Using the formula- Magnetic Potential energy $(u)=\frac{1}{2} \mathrm{Li}^{2}$ $=\frac{1}{2} 4 \times 10^{-6} \times 2^{2}$ $=8 \times 10^{-6} \mathrm{~J}$ $=8 \mu \mathrm{J}$
NEET (UG)-07.05.2023
Electro Magnetic Induction
154723
A $12 \mathrm{~V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{~ms}$. The emf induced across the coil is $20 \mathrm{~V}$. The inductance of the coil is:
1 $5 \mathrm{mH}$
2 $12 \mathrm{mH}$
3 $8 \mathrm{mH}$
4 $10 \mathrm{mH}$
Explanation:
D Given that, $\mathrm{E}=12 \mathrm{~V}$ Resistance, $\mathrm{R}=6 \Omega$ Induced emf, $\varepsilon=20 \mathrm{~V}$ Time, $\mathrm{t}=1 \mathrm{~ms}=1 \times 10^{-3} \mathrm{~s}$. Then, $i=\frac{E}{R}=\frac{12}{6}=2 \mathrm{~A}$ Thus, change in current $\left(\frac{\mathrm{di}}{\mathrm{dt}}\right)=-\frac{2}{1 \times 10^{-3}}=-2000 \mathrm{~A} / \mathrm{s}$ Where, negative sign (-ve) indicates the $\frac{\mathrm{di}}{\mathrm{dt}} \mathrm{is}$ decreasing. We know that, Induced emf $(\varepsilon)=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $20=-\mathrm{L} \times(-2000)$ $\mathrm{L}=10^{-2} \mathrm{H}$ $\mathrm{L}=10 \mathrm{mH}$
JEE Main-15.04.2023
Electro Magnetic Induction
154725
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' $R$ ' is placed inside a large square loop of wire of side $L(L>>R)$. The loops are coplanar and their centers coincide:
154726
A magnetic field of flux density $1.0 \mathrm{~Wb} \mathrm{~m}^{-2}$ acts normal to a 80 turn coil of $0.01 \mathrm{~m}^2$ area. If this coil is removed from the field in 0.2 second, the emf induced in it is
154728
A long solenoid has 500 turns. When a current of $2 \mathrm{~A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \mathrm{~Wb}$. The self-inductance of the solenoid is
1 $2.5 \mathrm{H}$
2 $2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $4 \mathrm{H}$
Explanation:
C Given that, Number of turns, $\mathrm{N}=500$ turn, Current, $\mathrm{I}=2 \mathrm{~A}$, Magnetic flux, $\phi=4 \times 10^{-3} \mathrm{~Wb}$ Self inductance, $L=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{500 \times 4 \times 10^{-3}}{2}=5 \times 2 \times 10^{-1}$ $\mathrm{~L}=1 \mathrm{H}$
154722
The magnetic energy stored in an inductor of inductance $4 \mu \mathrm{H}$ carrying a current of $2 \mathrm{~A}$ is:
1 $4 \mathrm{~mJ}$
2 $8 \mathrm{~mJ}$
3 $8 \mu \mathrm{J}$
4 $4 \mu \mathrm{J}$
Explanation:
C Given, Inductance $(\mathrm{L})=4 \mu \mathrm{H}$ Current (i) $=2 \mathrm{~A}$ Using the formula- Magnetic Potential energy $(u)=\frac{1}{2} \mathrm{Li}^{2}$ $=\frac{1}{2} 4 \times 10^{-6} \times 2^{2}$ $=8 \times 10^{-6} \mathrm{~J}$ $=8 \mu \mathrm{J}$
NEET (UG)-07.05.2023
Electro Magnetic Induction
154723
A $12 \mathrm{~V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{~ms}$. The emf induced across the coil is $20 \mathrm{~V}$. The inductance of the coil is:
1 $5 \mathrm{mH}$
2 $12 \mathrm{mH}$
3 $8 \mathrm{mH}$
4 $10 \mathrm{mH}$
Explanation:
D Given that, $\mathrm{E}=12 \mathrm{~V}$ Resistance, $\mathrm{R}=6 \Omega$ Induced emf, $\varepsilon=20 \mathrm{~V}$ Time, $\mathrm{t}=1 \mathrm{~ms}=1 \times 10^{-3} \mathrm{~s}$. Then, $i=\frac{E}{R}=\frac{12}{6}=2 \mathrm{~A}$ Thus, change in current $\left(\frac{\mathrm{di}}{\mathrm{dt}}\right)=-\frac{2}{1 \times 10^{-3}}=-2000 \mathrm{~A} / \mathrm{s}$ Where, negative sign (-ve) indicates the $\frac{\mathrm{di}}{\mathrm{dt}} \mathrm{is}$ decreasing. We know that, Induced emf $(\varepsilon)=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $20=-\mathrm{L} \times(-2000)$ $\mathrm{L}=10^{-2} \mathrm{H}$ $\mathrm{L}=10 \mathrm{mH}$
JEE Main-15.04.2023
Electro Magnetic Induction
154725
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' $R$ ' is placed inside a large square loop of wire of side $L(L>>R)$. The loops are coplanar and their centers coincide:
154726
A magnetic field of flux density $1.0 \mathrm{~Wb} \mathrm{~m}^{-2}$ acts normal to a 80 turn coil of $0.01 \mathrm{~m}^2$ area. If this coil is removed from the field in 0.2 second, the emf induced in it is
154728
A long solenoid has 500 turns. When a current of $2 \mathrm{~A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \mathrm{~Wb}$. The self-inductance of the solenoid is
1 $2.5 \mathrm{H}$
2 $2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $4 \mathrm{H}$
Explanation:
C Given that, Number of turns, $\mathrm{N}=500$ turn, Current, $\mathrm{I}=2 \mathrm{~A}$, Magnetic flux, $\phi=4 \times 10^{-3} \mathrm{~Wb}$ Self inductance, $L=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{500 \times 4 \times 10^{-3}}{2}=5 \times 2 \times 10^{-1}$ $\mathrm{~L}=1 \mathrm{H}$
154722
The magnetic energy stored in an inductor of inductance $4 \mu \mathrm{H}$ carrying a current of $2 \mathrm{~A}$ is:
1 $4 \mathrm{~mJ}$
2 $8 \mathrm{~mJ}$
3 $8 \mu \mathrm{J}$
4 $4 \mu \mathrm{J}$
Explanation:
C Given, Inductance $(\mathrm{L})=4 \mu \mathrm{H}$ Current (i) $=2 \mathrm{~A}$ Using the formula- Magnetic Potential energy $(u)=\frac{1}{2} \mathrm{Li}^{2}$ $=\frac{1}{2} 4 \times 10^{-6} \times 2^{2}$ $=8 \times 10^{-6} \mathrm{~J}$ $=8 \mu \mathrm{J}$
NEET (UG)-07.05.2023
Electro Magnetic Induction
154723
A $12 \mathrm{~V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{~ms}$. The emf induced across the coil is $20 \mathrm{~V}$. The inductance of the coil is:
1 $5 \mathrm{mH}$
2 $12 \mathrm{mH}$
3 $8 \mathrm{mH}$
4 $10 \mathrm{mH}$
Explanation:
D Given that, $\mathrm{E}=12 \mathrm{~V}$ Resistance, $\mathrm{R}=6 \Omega$ Induced emf, $\varepsilon=20 \mathrm{~V}$ Time, $\mathrm{t}=1 \mathrm{~ms}=1 \times 10^{-3} \mathrm{~s}$. Then, $i=\frac{E}{R}=\frac{12}{6}=2 \mathrm{~A}$ Thus, change in current $\left(\frac{\mathrm{di}}{\mathrm{dt}}\right)=-\frac{2}{1 \times 10^{-3}}=-2000 \mathrm{~A} / \mathrm{s}$ Where, negative sign (-ve) indicates the $\frac{\mathrm{di}}{\mathrm{dt}} \mathrm{is}$ decreasing. We know that, Induced emf $(\varepsilon)=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $20=-\mathrm{L} \times(-2000)$ $\mathrm{L}=10^{-2} \mathrm{H}$ $\mathrm{L}=10 \mathrm{mH}$
JEE Main-15.04.2023
Electro Magnetic Induction
154725
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' $R$ ' is placed inside a large square loop of wire of side $L(L>>R)$. The loops are coplanar and their centers coincide:
154726
A magnetic field of flux density $1.0 \mathrm{~Wb} \mathrm{~m}^{-2}$ acts normal to a 80 turn coil of $0.01 \mathrm{~m}^2$ area. If this coil is removed from the field in 0.2 second, the emf induced in it is
154728
A long solenoid has 500 turns. When a current of $2 \mathrm{~A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \mathrm{~Wb}$. The self-inductance of the solenoid is
1 $2.5 \mathrm{H}$
2 $2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $4 \mathrm{H}$
Explanation:
C Given that, Number of turns, $\mathrm{N}=500$ turn, Current, $\mathrm{I}=2 \mathrm{~A}$, Magnetic flux, $\phi=4 \times 10^{-3} \mathrm{~Wb}$ Self inductance, $L=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{500 \times 4 \times 10^{-3}}{2}=5 \times 2 \times 10^{-1}$ $\mathrm{~L}=1 \mathrm{H}$
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Electro Magnetic Induction
154722
The magnetic energy stored in an inductor of inductance $4 \mu \mathrm{H}$ carrying a current of $2 \mathrm{~A}$ is:
1 $4 \mathrm{~mJ}$
2 $8 \mathrm{~mJ}$
3 $8 \mu \mathrm{J}$
4 $4 \mu \mathrm{J}$
Explanation:
C Given, Inductance $(\mathrm{L})=4 \mu \mathrm{H}$ Current (i) $=2 \mathrm{~A}$ Using the formula- Magnetic Potential energy $(u)=\frac{1}{2} \mathrm{Li}^{2}$ $=\frac{1}{2} 4 \times 10^{-6} \times 2^{2}$ $=8 \times 10^{-6} \mathrm{~J}$ $=8 \mu \mathrm{J}$
NEET (UG)-07.05.2023
Electro Magnetic Induction
154723
A $12 \mathrm{~V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{~ms}$. The emf induced across the coil is $20 \mathrm{~V}$. The inductance of the coil is:
1 $5 \mathrm{mH}$
2 $12 \mathrm{mH}$
3 $8 \mathrm{mH}$
4 $10 \mathrm{mH}$
Explanation:
D Given that, $\mathrm{E}=12 \mathrm{~V}$ Resistance, $\mathrm{R}=6 \Omega$ Induced emf, $\varepsilon=20 \mathrm{~V}$ Time, $\mathrm{t}=1 \mathrm{~ms}=1 \times 10^{-3} \mathrm{~s}$. Then, $i=\frac{E}{R}=\frac{12}{6}=2 \mathrm{~A}$ Thus, change in current $\left(\frac{\mathrm{di}}{\mathrm{dt}}\right)=-\frac{2}{1 \times 10^{-3}}=-2000 \mathrm{~A} / \mathrm{s}$ Where, negative sign (-ve) indicates the $\frac{\mathrm{di}}{\mathrm{dt}} \mathrm{is}$ decreasing. We know that, Induced emf $(\varepsilon)=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $20=-\mathrm{L} \times(-2000)$ $\mathrm{L}=10^{-2} \mathrm{H}$ $\mathrm{L}=10 \mathrm{mH}$
JEE Main-15.04.2023
Electro Magnetic Induction
154725
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' $R$ ' is placed inside a large square loop of wire of side $L(L>>R)$. The loops are coplanar and their centers coincide:
154726
A magnetic field of flux density $1.0 \mathrm{~Wb} \mathrm{~m}^{-2}$ acts normal to a 80 turn coil of $0.01 \mathrm{~m}^2$ area. If this coil is removed from the field in 0.2 second, the emf induced in it is
154728
A long solenoid has 500 turns. When a current of $2 \mathrm{~A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \mathrm{~Wb}$. The self-inductance of the solenoid is
1 $2.5 \mathrm{H}$
2 $2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $4 \mathrm{H}$
Explanation:
C Given that, Number of turns, $\mathrm{N}=500$ turn, Current, $\mathrm{I}=2 \mathrm{~A}$, Magnetic flux, $\phi=4 \times 10^{-3} \mathrm{~Wb}$ Self inductance, $L=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{500 \times 4 \times 10^{-3}}{2}=5 \times 2 \times 10^{-1}$ $\mathrm{~L}=1 \mathrm{H}$
154722
The magnetic energy stored in an inductor of inductance $4 \mu \mathrm{H}$ carrying a current of $2 \mathrm{~A}$ is:
1 $4 \mathrm{~mJ}$
2 $8 \mathrm{~mJ}$
3 $8 \mu \mathrm{J}$
4 $4 \mu \mathrm{J}$
Explanation:
C Given, Inductance $(\mathrm{L})=4 \mu \mathrm{H}$ Current (i) $=2 \mathrm{~A}$ Using the formula- Magnetic Potential energy $(u)=\frac{1}{2} \mathrm{Li}^{2}$ $=\frac{1}{2} 4 \times 10^{-6} \times 2^{2}$ $=8 \times 10^{-6} \mathrm{~J}$ $=8 \mu \mathrm{J}$
NEET (UG)-07.05.2023
Electro Magnetic Induction
154723
A $12 \mathrm{~V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{~ms}$. The emf induced across the coil is $20 \mathrm{~V}$. The inductance of the coil is:
1 $5 \mathrm{mH}$
2 $12 \mathrm{mH}$
3 $8 \mathrm{mH}$
4 $10 \mathrm{mH}$
Explanation:
D Given that, $\mathrm{E}=12 \mathrm{~V}$ Resistance, $\mathrm{R}=6 \Omega$ Induced emf, $\varepsilon=20 \mathrm{~V}$ Time, $\mathrm{t}=1 \mathrm{~ms}=1 \times 10^{-3} \mathrm{~s}$. Then, $i=\frac{E}{R}=\frac{12}{6}=2 \mathrm{~A}$ Thus, change in current $\left(\frac{\mathrm{di}}{\mathrm{dt}}\right)=-\frac{2}{1 \times 10^{-3}}=-2000 \mathrm{~A} / \mathrm{s}$ Where, negative sign (-ve) indicates the $\frac{\mathrm{di}}{\mathrm{dt}} \mathrm{is}$ decreasing. We know that, Induced emf $(\varepsilon)=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $20=-\mathrm{L} \times(-2000)$ $\mathrm{L}=10^{-2} \mathrm{H}$ $\mathrm{L}=10 \mathrm{mH}$
JEE Main-15.04.2023
Electro Magnetic Induction
154725
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' $R$ ' is placed inside a large square loop of wire of side $L(L>>R)$. The loops are coplanar and their centers coincide:
154726
A magnetic field of flux density $1.0 \mathrm{~Wb} \mathrm{~m}^{-2}$ acts normal to a 80 turn coil of $0.01 \mathrm{~m}^2$ area. If this coil is removed from the field in 0.2 second, the emf induced in it is
154728
A long solenoid has 500 turns. When a current of $2 \mathrm{~A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \mathrm{~Wb}$. The self-inductance of the solenoid is
1 $2.5 \mathrm{H}$
2 $2 \mathrm{H}$
3 $1 \mathrm{H}$
4 $4 \mathrm{H}$
Explanation:
C Given that, Number of turns, $\mathrm{N}=500$ turn, Current, $\mathrm{I}=2 \mathrm{~A}$, Magnetic flux, $\phi=4 \times 10^{-3} \mathrm{~Wb}$ Self inductance, $L=\frac{\mathrm{N} \phi}{\mathrm{I}}$ $\mathrm{L}=\frac{500 \times 4 \times 10^{-3}}{2}=5 \times 2 \times 10^{-1}$ $\mathrm{~L}=1 \mathrm{H}$
