QBManager

Moving Charges & Magnetism

153419 A solenoid has a core of a material with relative permeability 400 . The winding $s$ of the solenoid are insulated from the core and carry a current of $1 A$ . If the number of turns is $1000$ per meter, find magnetic field (B)_ T. $(μ_{0} =$ $4 π \times 10^{- 7} SI$ )

1

1.6 π \times 10^{+ 2}

2

16 π \times 10^{2}

3

16 π \times 10^{- 2}

4

0.16 π \times 10^{- 2}

Explanation:

C Given that, $μ_{r} = 400, i = 1 A, n = 1000$ turn/metre
Then, $H = nI$
$= 1000 \times 1 = 1000 A / meter$
We know that, $B = μ H = μ_{0} μ_{r} H$
Then, $B = 4 π \times 10^{- 7} \times 400 \times 1000$
$B = 16 π \times 10^{- 2} T$

GUJCET 18.04.2022

Moving Charges & Magnetism

153422 A closely wound solenoid of length $1 m$ has 5 layers of 500 turns each. If the magnitude of magnetic field inside the solenoid near its centre is $4.4 mT$ , the current carried is-

1

1.4 A

2

1.5 A

3

1.6 A

4

1.8 A

Explanation:

A Given, length of solenoid $(l) = 1 m$
5 layer of winding of 500 turns then, total number turns on the solenoid $(N) = 5 \times 500 = 2500$
Magnetic field inside the solenoid $(B) = 4.4 mT =$ $4.4 \times 10^{- 3} T$
We know that,
$B = \frac{μ_{0} NI}{l}$
$I = \frac{B l}{μ_{o} N}$
$I = \frac{4.4 \times 10^{- 3} \times 1}{4 π \times 10^{- 7} \times 2500}$
$I = 1.4 A$

AP EAMCET-08.07.2022

Moving Charges & Magnetism

153423 The magnetic field at the center of current carrying circular loop is $B_{1}$ . The magnetic field at a distance of $\sqrt{3}$ times radius of the given circular loop from the center on its axis is $B_{2}$ . The value of $B_{1} / B_{2}$ will be

1

9 : 4

2

12 : \sqrt{5}

3

8 : 1

4

5 : \sqrt{3}

Explanation:

C Magnetic field at the centre of current carrying circular loop-
$B_{1} = \frac{μ_{0} I}{2 a}$
Where, $a =$ radius of circular loop
Magnetic field at a distance of $\sqrt{3}$ times radius-
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {(a^{2} + x^{2})}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {[a^{2} + (\sqrt{3} a)^{2}]}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 (8 a^{3})} = \frac{μ_{0} I}{16 a}$
ratio of equation (i) and (ii),
$∴ B_{1} = \frac{\frac{μ_{0} I}{2 a}}{\frac{μ_{0} I}{16 a}}$
$\frac{B_{1}}{B_{2}} = 8 : 1$

JEE Main-28.07.2022

Moving Charges & Magnetism

153424 A long solenoid has 70 turns/cm and carries current I. An electron moves within the solenoid in a circle of radius $2.5 cm$ perpendicular to the solenoid axis. If the speed of the electron is $4.4 \times 10^{6} m / s$ then the current $I$ in the solenoid is
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI unit, mass of electron $=$ $9 \times 10^{- 31} Kg$ , charge on electron $= 1.6 \times 10^{- 19} C$ )

1

98.5 mA

2

112.5 mA

3

125 mA

4

175.0 mA

Explanation:

B Given,
$n = 70 turns / cm = 7000 turns / m$
$r = 2.5 cm = 2.5 \times 10^{- 2} m$
$v = 4.4 \times 10^{6} m / s$
$μ_{0} = 4 π \times 10^{- 7}$
$m = 9 \times 10^{- 31} kg$
$e = 1.6 \times 10^{- 19} C$
We know that, an orbital radius for the electron,
$r = \frac{mv}{eB}, B = \frac{mv}{er}$
$μ_{0} nI = \frac{mv}{er}$
$I = \frac{mv}{e μ_{0} nr}$
$I = \frac{9 \times 10^{- 31} \times 4.4 \times 10^{6}}{1.6 \times 10^{- 19} \times 4 π \times 10^{- 7} \times 70 \times 2.5}$
$I = 0.1126$
$I = 112.6 \times 10^{- 3}$
$I \approx 112.5 mA$

TS EAMCET 18.07.2022

Moving Charges & Magnetism

153419 A solenoid has a core of a material with relative permeability 400 . The winding $s$ of the solenoid are insulated from the core and carry a current of $1 A$ . If the number of turns is $1000$ per meter, find magnetic field (B)_ T. $(μ_{0} =$ $4 π \times 10^{- 7} SI$ )

1

1.6 π \times 10^{+ 2}

2

16 π \times 10^{2}

3

16 π \times 10^{- 2}

4

0.16 π \times 10^{- 2}

Explanation:

C Given that, $μ_{r} = 400, i = 1 A, n = 1000$ turn/metre
Then, $H = nI$
$= 1000 \times 1 = 1000 A / meter$
We know that, $B = μ H = μ_{0} μ_{r} H$
Then, $B = 4 π \times 10^{- 7} \times 400 \times 1000$
$B = 16 π \times 10^{- 2} T$

GUJCET 18.04.2022

Moving Charges & Magnetism

153422 A closely wound solenoid of length $1 m$ has 5 layers of 500 turns each. If the magnitude of magnetic field inside the solenoid near its centre is $4.4 mT$ , the current carried is-

1

1.4 A

2

1.5 A

3

1.6 A

4

1.8 A

Explanation:

A Given, length of solenoid $(l) = 1 m$
5 layer of winding of 500 turns then, total number turns on the solenoid $(N) = 5 \times 500 = 2500$
Magnetic field inside the solenoid $(B) = 4.4 mT =$ $4.4 \times 10^{- 3} T$
We know that,
$B = \frac{μ_{0} NI}{l}$
$I = \frac{B l}{μ_{o} N}$
$I = \frac{4.4 \times 10^{- 3} \times 1}{4 π \times 10^{- 7} \times 2500}$
$I = 1.4 A$

AP EAMCET-08.07.2022

Moving Charges & Magnetism

153423 The magnetic field at the center of current carrying circular loop is $B_{1}$ . The magnetic field at a distance of $\sqrt{3}$ times radius of the given circular loop from the center on its axis is $B_{2}$ . The value of $B_{1} / B_{2}$ will be

1

9 : 4

2

12 : \sqrt{5}

3

8 : 1

4

5 : \sqrt{3}

Explanation:

C Magnetic field at the centre of current carrying circular loop-
$B_{1} = \frac{μ_{0} I}{2 a}$
Where, $a =$ radius of circular loop
Magnetic field at a distance of $\sqrt{3}$ times radius-
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {(a^{2} + x^{2})}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {[a^{2} + (\sqrt{3} a)^{2}]}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 (8 a^{3})} = \frac{μ_{0} I}{16 a}$
ratio of equation (i) and (ii),
$∴ B_{1} = \frac{\frac{μ_{0} I}{2 a}}{\frac{μ_{0} I}{16 a}}$
$\frac{B_{1}}{B_{2}} = 8 : 1$

JEE Main-28.07.2022

Moving Charges & Magnetism

153424 A long solenoid has 70 turns/cm and carries current I. An electron moves within the solenoid in a circle of radius $2.5 cm$ perpendicular to the solenoid axis. If the speed of the electron is $4.4 \times 10^{6} m / s$ then the current $I$ in the solenoid is
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI unit, mass of electron $=$ $9 \times 10^{- 31} Kg$ , charge on electron $= 1.6 \times 10^{- 19} C$ )

1

98.5 mA

2

112.5 mA

3

125 mA

4

175.0 mA

Explanation:

B Given,
$n = 70 turns / cm = 7000 turns / m$
$r = 2.5 cm = 2.5 \times 10^{- 2} m$
$v = 4.4 \times 10^{6} m / s$
$μ_{0} = 4 π \times 10^{- 7}$
$m = 9 \times 10^{- 31} kg$
$e = 1.6 \times 10^{- 19} C$
We know that, an orbital radius for the electron,
$r = \frac{mv}{eB}, B = \frac{mv}{er}$
$μ_{0} nI = \frac{mv}{er}$
$I = \frac{mv}{e μ_{0} nr}$
$I = \frac{9 \times 10^{- 31} \times 4.4 \times 10^{6}}{1.6 \times 10^{- 19} \times 4 π \times 10^{- 7} \times 70 \times 2.5}$
$I = 0.1126$
$I = 112.6 \times 10^{- 3}$
$I \approx 112.5 mA$

TS EAMCET 18.07.2022

Moving Charges & Magnetism

153419 A solenoid has a core of a material with relative permeability 400 . The winding $s$ of the solenoid are insulated from the core and carry a current of $1 A$ . If the number of turns is $1000$ per meter, find magnetic field (B)_ T. $(μ_{0} =$ $4 π \times 10^{- 7} SI$ )

1

1.6 π \times 10^{+ 2}

2

16 π \times 10^{2}

3

16 π \times 10^{- 2}

4

0.16 π \times 10^{- 2}

Explanation:

C Given that, $μ_{r} = 400, i = 1 A, n = 1000$ turn/metre
Then, $H = nI$
$= 1000 \times 1 = 1000 A / meter$
We know that, $B = μ H = μ_{0} μ_{r} H$
Then, $B = 4 π \times 10^{- 7} \times 400 \times 1000$
$B = 16 π \times 10^{- 2} T$

GUJCET 18.04.2022

Moving Charges & Magnetism

153422 A closely wound solenoid of length $1 m$ has 5 layers of 500 turns each. If the magnitude of magnetic field inside the solenoid near its centre is $4.4 mT$ , the current carried is-

1

1.4 A

2

1.5 A

3

1.6 A

4

1.8 A

Explanation:

A Given, length of solenoid $(l) = 1 m$
5 layer of winding of 500 turns then, total number turns on the solenoid $(N) = 5 \times 500 = 2500$
Magnetic field inside the solenoid $(B) = 4.4 mT =$ $4.4 \times 10^{- 3} T$
We know that,
$B = \frac{μ_{0} NI}{l}$
$I = \frac{B l}{μ_{o} N}$
$I = \frac{4.4 \times 10^{- 3} \times 1}{4 π \times 10^{- 7} \times 2500}$
$I = 1.4 A$

AP EAMCET-08.07.2022

Moving Charges & Magnetism

153423 The magnetic field at the center of current carrying circular loop is $B_{1}$ . The magnetic field at a distance of $\sqrt{3}$ times radius of the given circular loop from the center on its axis is $B_{2}$ . The value of $B_{1} / B_{2}$ will be

1

9 : 4

2

12 : \sqrt{5}

3

8 : 1

4

5 : \sqrt{3}

Explanation:

C Magnetic field at the centre of current carrying circular loop-
$B_{1} = \frac{μ_{0} I}{2 a}$
Where, $a =$ radius of circular loop
Magnetic field at a distance of $\sqrt{3}$ times radius-
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {(a^{2} + x^{2})}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {[a^{2} + (\sqrt{3} a)^{2}]}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 (8 a^{3})} = \frac{μ_{0} I}{16 a}$
ratio of equation (i) and (ii),
$∴ B_{1} = \frac{\frac{μ_{0} I}{2 a}}{\frac{μ_{0} I}{16 a}}$
$\frac{B_{1}}{B_{2}} = 8 : 1$

JEE Main-28.07.2022

Moving Charges & Magnetism

153424 A long solenoid has 70 turns/cm and carries current I. An electron moves within the solenoid in a circle of radius $2.5 cm$ perpendicular to the solenoid axis. If the speed of the electron is $4.4 \times 10^{6} m / s$ then the current $I$ in the solenoid is
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI unit, mass of electron $=$ $9 \times 10^{- 31} Kg$ , charge on electron $= 1.6 \times 10^{- 19} C$ )

1

98.5 mA

2

112.5 mA

3

125 mA

4

175.0 mA

Explanation:

B Given,
$n = 70 turns / cm = 7000 turns / m$
$r = 2.5 cm = 2.5 \times 10^{- 2} m$
$v = 4.4 \times 10^{6} m / s$
$μ_{0} = 4 π \times 10^{- 7}$
$m = 9 \times 10^{- 31} kg$
$e = 1.6 \times 10^{- 19} C$
We know that, an orbital radius for the electron,
$r = \frac{mv}{eB}, B = \frac{mv}{er}$
$μ_{0} nI = \frac{mv}{er}$
$I = \frac{mv}{e μ_{0} nr}$
$I = \frac{9 \times 10^{- 31} \times 4.4 \times 10^{6}}{1.6 \times 10^{- 19} \times 4 π \times 10^{- 7} \times 70 \times 2.5}$
$I = 0.1126$
$I = 112.6 \times 10^{- 3}$
$I \approx 112.5 mA$

TS EAMCET 18.07.2022

Moving Charges & Magnetism

153419 A solenoid has a core of a material with relative permeability 400 . The winding $s$ of the solenoid are insulated from the core and carry a current of $1 A$ . If the number of turns is $1000$ per meter, find magnetic field (B)_ T. $(μ_{0} =$ $4 π \times 10^{- 7} SI$ )

1

1.6 π \times 10^{+ 2}

2

16 π \times 10^{2}

3

16 π \times 10^{- 2}

4

0.16 π \times 10^{- 2}

Explanation:

C Given that, $μ_{r} = 400, i = 1 A, n = 1000$ turn/metre
Then, $H = nI$
$= 1000 \times 1 = 1000 A / meter$
We know that, $B = μ H = μ_{0} μ_{r} H$
Then, $B = 4 π \times 10^{- 7} \times 400 \times 1000$
$B = 16 π \times 10^{- 2} T$

GUJCET 18.04.2022

Moving Charges & Magnetism

153422 A closely wound solenoid of length $1 m$ has 5 layers of 500 turns each. If the magnitude of magnetic field inside the solenoid near its centre is $4.4 mT$ , the current carried is-

1

1.4 A

2

1.5 A

3

1.6 A

4

1.8 A

Explanation:

A Given, length of solenoid $(l) = 1 m$
5 layer of winding of 500 turns then, total number turns on the solenoid $(N) = 5 \times 500 = 2500$
Magnetic field inside the solenoid $(B) = 4.4 mT =$ $4.4 \times 10^{- 3} T$
We know that,
$B = \frac{μ_{0} NI}{l}$
$I = \frac{B l}{μ_{o} N}$
$I = \frac{4.4 \times 10^{- 3} \times 1}{4 π \times 10^{- 7} \times 2500}$
$I = 1.4 A$

AP EAMCET-08.07.2022

Moving Charges & Magnetism

153423 The magnetic field at the center of current carrying circular loop is $B_{1}$ . The magnetic field at a distance of $\sqrt{3}$ times radius of the given circular loop from the center on its axis is $B_{2}$ . The value of $B_{1} / B_{2}$ will be

1

9 : 4

2

12 : \sqrt{5}

3

8 : 1

4

5 : \sqrt{3}

Explanation:

C Magnetic field at the centre of current carrying circular loop-
$B_{1} = \frac{μ_{0} I}{2 a}$
Where, $a =$ radius of circular loop
Magnetic field at a distance of $\sqrt{3}$ times radius-
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {(a^{2} + x^{2})}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 {[a^{2} + (\sqrt{3} a)^{2}]}^{3 / 2}}$
$B_{2} = \frac{μ_{0} {Ia}^{2}}{2 (8 a^{3})} = \frac{μ_{0} I}{16 a}$
ratio of equation (i) and (ii),
$∴ B_{1} = \frac{\frac{μ_{0} I}{2 a}}{\frac{μ_{0} I}{16 a}}$
$\frac{B_{1}}{B_{2}} = 8 : 1$

JEE Main-28.07.2022

Moving Charges & Magnetism

153424 A long solenoid has 70 turns/cm and carries current I. An electron moves within the solenoid in a circle of radius $2.5 cm$ perpendicular to the solenoid axis. If the speed of the electron is $4.4 \times 10^{6} m / s$ then the current $I$ in the solenoid is
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI unit, mass of electron $=$ $9 \times 10^{- 31} Kg$ , charge on electron $= 1.6 \times 10^{- 19} C$ )

1

98.5 mA

2

112.5 mA

3

125 mA

4

175.0 mA

Explanation:

B Given,
$n = 70 turns / cm = 7000 turns / m$
$r = 2.5 cm = 2.5 \times 10^{- 2} m$
$v = 4.4 \times 10^{6} m / s$
$μ_{0} = 4 π \times 10^{- 7}$
$m = 9 \times 10^{- 31} kg$
$e = 1.6 \times 10^{- 19} C$
We know that, an orbital radius for the electron,
$r = \frac{mv}{eB}, B = \frac{mv}{er}$
$μ_{0} nI = \frac{mv}{er}$
$I = \frac{mv}{e μ_{0} nr}$
$I = \frac{9 \times 10^{- 31} \times 4.4 \times 10^{6}}{1.6 \times 10^{- 19} \times 4 π \times 10^{- 7} \times 70 \times 2.5}$
$I = 0.1126$
$I = 112.6 \times 10^{- 3}$
$I \approx 112.5 mA$

TS EAMCET 18.07.2022

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: