QBManager

PHXII04:MOVING CHARGES AND MAGNETISM

362626 A thin hemispherical shell having a uniform surface charge density $σ$ spins with an angular speed $ω$ . If the magnetic field at $O$ is found to be $\frac{N μ_{0} σ ω}{3}$ , find the value of $N$ .
supporting img

1

\frac{2 μ_{0} σ ω}{3}

2

\frac{5 μ_{0} σ ω}{2}

3

\frac{4 μ_{0} σ ω}{1}

4

\frac{7 μ_{0} σ ω}{5}

Explanation:

The magnetic field due to the ring is $d B = \frac{μ_{0} (d i) r^{2}}{2 R^{3}}$
supporting img
The total magnetic field is
$B = \int d B = \frac{μ_{0}}{2 R^{3}} \int r^{2} d i$ ,
where $d i = (2 π r \cdot R d θ σ) \frac{ω}{2 π}$
or $B = \frac{μ_{0}}{2 R^{3}} \int r^{2} (2 π r R d θ σ) \frac{ω}{2 π}$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int r^{3} d θ$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int_{0}^{π / 2} R^{3} \sin^{3} θ d θ$
Evaluating the integration, we have $B = \frac{2 μ_{0} σ ω}{3}$

PHXII04:MOVING CHARGES AND MAGNETISM

362627 In the hydrogen atom, the electron is making $6.6 \times 10^{15} r p s$ . If the radius of the orbit is $0.53 \times 10^{- 10} m$ , then magnetic field produced at the centre of the orbit is

1

140 T

2

12.5 T

3

1.4 T

4

0.14 T

Explanation:

Here, $v = 6.6 \times 10^{15} r p s = 6.6 \times 10^{15} H z$
$q = e = 1.6 \times 10^{- 19} C, r = 0.53 \times 10^{- 10} m$
Magnetic field, $B = \frac{μ_{0} I}{2 r} = \frac{μ_{0} q v}{2 r} (∵ I = q v)$
$= \frac{4 π \times 10^{- 7} \times 1.6 \times 10^{- 19} \times 6.6 \times 10^{15}}{2 \times 0.53 \times 10^{- 10}}$
$= 12.5 T$

PHXII04:MOVING CHARGES AND MAGNETISM

362628 Charge $q$ is uniformly spread on a thin ring of radius $R$ . The ring rotates about its axis with a uniform frequency $f H z$ . The magnitude of magnetic induction at the centre of the ring is

1

\frac{μ_{0} q f}{2 R}

2

\frac{μ_{0} q}{2 f R}

3

\frac{μ_{0} q}{2 π f R}

4

\frac{μ_{0} q f}{2 π R}

Explanation:

The current in the ring is
$1 = q f (1)$
$The magnitude of magnetic induction at the centre of the ring is
$B = \frac{μ_{0} I}{2 R} = \frac{μ_{0} q f}{2 R} (using (1))$

PHXII04:MOVING CHARGES AND MAGNETISM

362629 A charge of 1 $C$ is placed at one end of a conducting rod of length 0.6 $m$ . The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency $10^{4} π {rads}^{- 1}$ . The magnetic field at a point on the axis of rotation at a distance of $0.8 m$ from the centre of the path is $B_{1}$ . Now, half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the midpoint of the rod with same angular frequency. The magnetic field at a point on the axis at a distance of 0.4 $m$ from the centre of the rod is $B_{2}$ then $\frac{B_{2}}{B_{1}} =$

1 4

2 8

3 2

4

2^{1 / 3}

Explanation:

A revolving charge is equivalent to a current,
$i_{1} = q . f = q \cdot \frac{ω}{2 π} = 1 \times \frac{10^{4}}{2} = 5 \times 10^{3} A$
The field at a distance $z$ from the centre of the axis of a current carrying coil is given by
supporting img
$B_{1} = \frac{μ_{0} a^{2} i_{1}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$= \frac{4 π \times 10^{- 7} \times 5 \times 10^{3} (0.6)^{2}}{2 {[(0.6)^{2} + (0.8)^{2}]}^{3 / 2}} = 1.13 \times 10^{- 3} T$
Equivalent current,
$i_{2} = \frac{q}{2} f + \frac{q}{2} f = q f = 5 \times 10^{3} A$
Magnetic field at a point $P$ in this case
supporting img
$B_{2} = \frac{μ_{0} a^{2} i_{2}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$B_{2} = \frac{4 π \times 10^{- 7} \times {(0.3)}^{2} \times 5 \times 10^{3}}{2 {({(0.3)}^{2} + {(0.4)}^{2})}^{3 / 2}}$
Here, $a = 0.3, z = 0.4 m$
which gives, $B_{2} = 2.26 \times 10^{- 3} T$

PHXII04:MOVING CHARGES AND MAGNETISM

362626 A thin hemispherical shell having a uniform surface charge density $σ$ spins with an angular speed $ω$ . If the magnetic field at $O$ is found to be $\frac{N μ_{0} σ ω}{3}$ , find the value of $N$ .
supporting img

1

\frac{2 μ_{0} σ ω}{3}

2

\frac{5 μ_{0} σ ω}{2}

3

\frac{4 μ_{0} σ ω}{1}

4

\frac{7 μ_{0} σ ω}{5}

Explanation:

The magnetic field due to the ring is $d B = \frac{μ_{0} (d i) r^{2}}{2 R^{3}}$
supporting img
The total magnetic field is
$B = \int d B = \frac{μ_{0}}{2 R^{3}} \int r^{2} d i$ ,
where $d i = (2 π r \cdot R d θ σ) \frac{ω}{2 π}$
or $B = \frac{μ_{0}}{2 R^{3}} \int r^{2} (2 π r R d θ σ) \frac{ω}{2 π}$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int r^{3} d θ$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int_{0}^{π / 2} R^{3} \sin^{3} θ d θ$
Evaluating the integration, we have $B = \frac{2 μ_{0} σ ω}{3}$

PHXII04:MOVING CHARGES AND MAGNETISM

362627 In the hydrogen atom, the electron is making $6.6 \times 10^{15} r p s$ . If the radius of the orbit is $0.53 \times 10^{- 10} m$ , then magnetic field produced at the centre of the orbit is

1

140 T

2

12.5 T

3

1.4 T

4

0.14 T

Explanation:

Here, $v = 6.6 \times 10^{15} r p s = 6.6 \times 10^{15} H z$
$q = e = 1.6 \times 10^{- 19} C, r = 0.53 \times 10^{- 10} m$
Magnetic field, $B = \frac{μ_{0} I}{2 r} = \frac{μ_{0} q v}{2 r} (∵ I = q v)$
$= \frac{4 π \times 10^{- 7} \times 1.6 \times 10^{- 19} \times 6.6 \times 10^{15}}{2 \times 0.53 \times 10^{- 10}}$
$= 12.5 T$

PHXII04:MOVING CHARGES AND MAGNETISM

362628 Charge $q$ is uniformly spread on a thin ring of radius $R$ . The ring rotates about its axis with a uniform frequency $f H z$ . The magnitude of magnetic induction at the centre of the ring is

1

\frac{μ_{0} q f}{2 R}

2

\frac{μ_{0} q}{2 f R}

3

\frac{μ_{0} q}{2 π f R}

4

\frac{μ_{0} q f}{2 π R}

Explanation:

The current in the ring is
$1 = q f (1)$
$The magnitude of magnetic induction at the centre of the ring is
$B = \frac{μ_{0} I}{2 R} = \frac{μ_{0} q f}{2 R} (using (1))$

PHXII04:MOVING CHARGES AND MAGNETISM

362629 A charge of 1 $C$ is placed at one end of a conducting rod of length 0.6 $m$ . The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency $10^{4} π {rads}^{- 1}$ . The magnetic field at a point on the axis of rotation at a distance of $0.8 m$ from the centre of the path is $B_{1}$ . Now, half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the midpoint of the rod with same angular frequency. The magnetic field at a point on the axis at a distance of 0.4 $m$ from the centre of the rod is $B_{2}$ then $\frac{B_{2}}{B_{1}} =$

1 4

2 8

3 2

4

2^{1 / 3}

Explanation:

A revolving charge is equivalent to a current,
$i_{1} = q . f = q \cdot \frac{ω}{2 π} = 1 \times \frac{10^{4}}{2} = 5 \times 10^{3} A$
The field at a distance $z$ from the centre of the axis of a current carrying coil is given by
supporting img
$B_{1} = \frac{μ_{0} a^{2} i_{1}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$= \frac{4 π \times 10^{- 7} \times 5 \times 10^{3} (0.6)^{2}}{2 {[(0.6)^{2} + (0.8)^{2}]}^{3 / 2}} = 1.13 \times 10^{- 3} T$
Equivalent current,
$i_{2} = \frac{q}{2} f + \frac{q}{2} f = q f = 5 \times 10^{3} A$
Magnetic field at a point $P$ in this case
supporting img
$B_{2} = \frac{μ_{0} a^{2} i_{2}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$B_{2} = \frac{4 π \times 10^{- 7} \times {(0.3)}^{2} \times 5 \times 10^{3}}{2 {({(0.3)}^{2} + {(0.4)}^{2})}^{3 / 2}}$
Here, $a = 0.3, z = 0.4 m$
which gives, $B_{2} = 2.26 \times 10^{- 3} T$

PHXII04:MOVING CHARGES AND MAGNETISM

362626 A thin hemispherical shell having a uniform surface charge density $σ$ spins with an angular speed $ω$ . If the magnetic field at $O$ is found to be $\frac{N μ_{0} σ ω}{3}$ , find the value of $N$ .
supporting img

1

\frac{2 μ_{0} σ ω}{3}

2

\frac{5 μ_{0} σ ω}{2}

3

\frac{4 μ_{0} σ ω}{1}

4

\frac{7 μ_{0} σ ω}{5}

Explanation:

The magnetic field due to the ring is $d B = \frac{μ_{0} (d i) r^{2}}{2 R^{3}}$
supporting img
The total magnetic field is
$B = \int d B = \frac{μ_{0}}{2 R^{3}} \int r^{2} d i$ ,
where $d i = (2 π r \cdot R d θ σ) \frac{ω}{2 π}$
or $B = \frac{μ_{0}}{2 R^{3}} \int r^{2} (2 π r R d θ σ) \frac{ω}{2 π}$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int r^{3} d θ$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int_{0}^{π / 2} R^{3} \sin^{3} θ d θ$
Evaluating the integration, we have $B = \frac{2 μ_{0} σ ω}{3}$

PHXII04:MOVING CHARGES AND MAGNETISM

362627 In the hydrogen atom, the electron is making $6.6 \times 10^{15} r p s$ . If the radius of the orbit is $0.53 \times 10^{- 10} m$ , then magnetic field produced at the centre of the orbit is

1

140 T

2

12.5 T

3

1.4 T

4

0.14 T

Explanation:

Here, $v = 6.6 \times 10^{15} r p s = 6.6 \times 10^{15} H z$
$q = e = 1.6 \times 10^{- 19} C, r = 0.53 \times 10^{- 10} m$
Magnetic field, $B = \frac{μ_{0} I}{2 r} = \frac{μ_{0} q v}{2 r} (∵ I = q v)$
$= \frac{4 π \times 10^{- 7} \times 1.6 \times 10^{- 19} \times 6.6 \times 10^{15}}{2 \times 0.53 \times 10^{- 10}}$
$= 12.5 T$

PHXII04:MOVING CHARGES AND MAGNETISM

362628 Charge $q$ is uniformly spread on a thin ring of radius $R$ . The ring rotates about its axis with a uniform frequency $f H z$ . The magnitude of magnetic induction at the centre of the ring is

1

\frac{μ_{0} q f}{2 R}

2

\frac{μ_{0} q}{2 f R}

3

\frac{μ_{0} q}{2 π f R}

4

\frac{μ_{0} q f}{2 π R}

Explanation:

The current in the ring is
$1 = q f (1)$
$The magnitude of magnetic induction at the centre of the ring is
$B = \frac{μ_{0} I}{2 R} = \frac{μ_{0} q f}{2 R} (using (1))$

PHXII04:MOVING CHARGES AND MAGNETISM

362629 A charge of 1 $C$ is placed at one end of a conducting rod of length 0.6 $m$ . The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency $10^{4} π {rads}^{- 1}$ . The magnetic field at a point on the axis of rotation at a distance of $0.8 m$ from the centre of the path is $B_{1}$ . Now, half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the midpoint of the rod with same angular frequency. The magnetic field at a point on the axis at a distance of 0.4 $m$ from the centre of the rod is $B_{2}$ then $\frac{B_{2}}{B_{1}} =$

1 4

2 8

3 2

4

2^{1 / 3}

Explanation:

A revolving charge is equivalent to a current,
$i_{1} = q . f = q \cdot \frac{ω}{2 π} = 1 \times \frac{10^{4}}{2} = 5 \times 10^{3} A$
The field at a distance $z$ from the centre of the axis of a current carrying coil is given by
supporting img
$B_{1} = \frac{μ_{0} a^{2} i_{1}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$= \frac{4 π \times 10^{- 7} \times 5 \times 10^{3} (0.6)^{2}}{2 {[(0.6)^{2} + (0.8)^{2}]}^{3 / 2}} = 1.13 \times 10^{- 3} T$
Equivalent current,
$i_{2} = \frac{q}{2} f + \frac{q}{2} f = q f = 5 \times 10^{3} A$
Magnetic field at a point $P$ in this case
supporting img
$B_{2} = \frac{μ_{0} a^{2} i_{2}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$B_{2} = \frac{4 π \times 10^{- 7} \times {(0.3)}^{2} \times 5 \times 10^{3}}{2 {({(0.3)}^{2} + {(0.4)}^{2})}^{3 / 2}}$
Here, $a = 0.3, z = 0.4 m$
which gives, $B_{2} = 2.26 \times 10^{- 3} T$

PHXII04:MOVING CHARGES AND MAGNETISM

362626 A thin hemispherical shell having a uniform surface charge density $σ$ spins with an angular speed $ω$ . If the magnetic field at $O$ is found to be $\frac{N μ_{0} σ ω}{3}$ , find the value of $N$ .
supporting img

1

\frac{2 μ_{0} σ ω}{3}

2

\frac{5 μ_{0} σ ω}{2}

3

\frac{4 μ_{0} σ ω}{1}

4

\frac{7 μ_{0} σ ω}{5}

Explanation:

The magnetic field due to the ring is $d B = \frac{μ_{0} (d i) r^{2}}{2 R^{3}}$
supporting img
The total magnetic field is
$B = \int d B = \frac{μ_{0}}{2 R^{3}} \int r^{2} d i$ ,
where $d i = (2 π r \cdot R d θ σ) \frac{ω}{2 π}$
or $B = \frac{μ_{0}}{2 R^{3}} \int r^{2} (2 π r R d θ σ) \frac{ω}{2 π}$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int r^{3} d θ$
$= \frac{π μ_{0} σ ω}{2 R^{2}} \int_{0}^{π / 2} R^{3} \sin^{3} θ d θ$
Evaluating the integration, we have $B = \frac{2 μ_{0} σ ω}{3}$

PHXII04:MOVING CHARGES AND MAGNETISM

362627 In the hydrogen atom, the electron is making $6.6 \times 10^{15} r p s$ . If the radius of the orbit is $0.53 \times 10^{- 10} m$ , then magnetic field produced at the centre of the orbit is

1

140 T

2

12.5 T

3

1.4 T

4

0.14 T

Explanation:

Here, $v = 6.6 \times 10^{15} r p s = 6.6 \times 10^{15} H z$
$q = e = 1.6 \times 10^{- 19} C, r = 0.53 \times 10^{- 10} m$
Magnetic field, $B = \frac{μ_{0} I}{2 r} = \frac{μ_{0} q v}{2 r} (∵ I = q v)$
$= \frac{4 π \times 10^{- 7} \times 1.6 \times 10^{- 19} \times 6.6 \times 10^{15}}{2 \times 0.53 \times 10^{- 10}}$
$= 12.5 T$

PHXII04:MOVING CHARGES AND MAGNETISM

362628 Charge $q$ is uniformly spread on a thin ring of radius $R$ . The ring rotates about its axis with a uniform frequency $f H z$ . The magnitude of magnetic induction at the centre of the ring is

1

\frac{μ_{0} q f}{2 R}

2

\frac{μ_{0} q}{2 f R}

3

\frac{μ_{0} q}{2 π f R}

4

\frac{μ_{0} q f}{2 π R}

Explanation:

The current in the ring is
$1 = q f (1)$
$The magnitude of magnetic induction at the centre of the ring is
$B = \frac{μ_{0} I}{2 R} = \frac{μ_{0} q f}{2 R} (using (1))$

PHXII04:MOVING CHARGES AND MAGNETISM

362629 A charge of 1 $C$ is placed at one end of a conducting rod of length 0.6 $m$ . The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency $10^{4} π {rads}^{- 1}$ . The magnetic field at a point on the axis of rotation at a distance of $0.8 m$ from the centre of the path is $B_{1}$ . Now, half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about a horizontal axis passing through the midpoint of the rod with same angular frequency. The magnetic field at a point on the axis at a distance of 0.4 $m$ from the centre of the rod is $B_{2}$ then $\frac{B_{2}}{B_{1}} =$

1 4

2 8

3 2

4

2^{1 / 3}

Explanation:

A revolving charge is equivalent to a current,
$i_{1} = q . f = q \cdot \frac{ω}{2 π} = 1 \times \frac{10^{4}}{2} = 5 \times 10^{3} A$
The field at a distance $z$ from the centre of the axis of a current carrying coil is given by
supporting img
$B_{1} = \frac{μ_{0} a^{2} i_{1}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$= \frac{4 π \times 10^{- 7} \times 5 \times 10^{3} (0.6)^{2}}{2 {[(0.6)^{2} + (0.8)^{2}]}^{3 / 2}} = 1.13 \times 10^{- 3} T$
Equivalent current,
$i_{2} = \frac{q}{2} f + \frac{q}{2} f = q f = 5 \times 10^{3} A$
Magnetic field at a point $P$ in this case
supporting img
$B_{2} = \frac{μ_{0} a^{2} i_{2}}{2 {(a^{2} + z^{2})}^{3 / 2}}$
$B_{2} = \frac{4 π \times 10^{- 7} \times {(0.3)}^{2} \times 5 \times 10^{3}}{2 {({(0.3)}^{2} + {(0.4)}^{2})}^{3 / 2}}$
Here, $a = 0.3, z = 0.4 m$
which gives, $B_{2} = 2.26 \times 10^{- 3} T$