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Moving Charges & Magnetism

153118 A bar magnet with a magnetic moment $5.0 {Am}^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T$ . The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

1 J

3

2 J

4 Zero

Explanation:

A Given,
$Magnetic moment (M) = 5 {Am}^{2}$
$Magnetic field (B) = 0.4 T$
We know that,
$W = \int τ . d θ$
$W = \int_{0}^{π} MB \sin θ d θ$
$W = MB \times [- \cos]_{0}^{π}$
$W = - 0.4 \times 5 [\cos 180^{\circ} - \cos 0^{\circ}]$
$W = - 0.4 \times 5 \times (- 2)$
$W = 4 J$

JEE Main-31.01.2023

Moving Charges & Magnetism

153119 A uniform current is flowing along the length of an infinite, straight, thin, hollow cylinder of radius $R$ . The magnetic field $B$ produced at a perpendicular distance $d$ from the axis of the cylinder is plotted in a graph. Which of the following figures looks like the plot?

1

2

3

4

Explanation:

C Let radius of hollow cylinder is $R$

For $d < R, B \times 2 π d = μ_{0} I_{in}$
$(∵$ Magnetic field inside conductor is zero)
$B = 0$
$(I_{in} = 0)$
For $d > R B \times 2 π d = μ_{0} I$
$B = \frac{μ_{0} I}{2 π d} \Rightarrow B \propto \frac{1}{d}$

JEE Main-28.06.2022

Moving Charges & Magnetism

153121 A short bar magnet produces a magnetic field of $6.4 \times 10^{- 5} T$ at a distance of $20 cm$ from the center of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 cm$ from the center of the magnet on the axis, is

1

4.8 \times 10^{- 5} T

2

3.2 \times 10^{- 5} T

3

1.6 \times 10^{- 5} T

4

6.4 \times 10^{- 5} T

Explanation:

C Given,
Normal bisector distance $(d_{1}) = 20 cm = 0.2 m$
Axial distance $(d_{2}) = 40 cm = 0.4 m$
Magnetic field at its bisector
$B_{bisector} = \frac{μ_{0}}{4 π} \times \frac{M}{d_{1}^{3}}$
Magnetic field at its axis
$B_{axis} = \frac{μ_{0}}{4 π} \times \frac{2 M}{d_{2}^{3}}$
Ratio of magnetic fields,
$\frac{B_{axis}}{B_{bisector}} = \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = B_{bisector} \times \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = 6.4 \times 10^{- 5} \times \frac{2 \times (0.2)^{3}}{(0.4)^{3}}$
$B_{axis} = 1.6 \times 10^{- 5} T$

AP EAMCET-07.07.2022

Moving Charges & Magnetism

153122 The magnetic field due to current carrying circular loop of radius $6 cm$ at a point on the axis at a distance of $8 cm$ from centre is $27 μ T$ . The magnetic field at the centre of the current carrying loop is.

1

75 μ T

.

2

125 μ T

3

150 μ T

.

4

250 μ T

.

Explanation:

B Given,
Circular loop of radius $(r) = 6 cm = 6 \times 10^{- 2} m$
Axial distance $(x) = 8 cm = 8 \times 10^{- 2} m$
Magnetic field at centre $= 27 μ T = 27 \times 10^{- 6} T$
$B = \frac{μ_{o}}{4 π} I 2 π \frac{r^{2}}{{(r^{2} + x^{2})}^{3 / 2}}$
$27 \times 10^{- 6} = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{I \times 2 π \times {(6 \times 10^{- 2})}^{2}}{{({(6 \times 10^{- 2})}^{2} + {(8 \times 10^{- 2})}^{2})}^{3 / 2}}$
$I = \frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}}$
Magnetic field at the centre
$B = \frac{μ_{0} I}{4 π r} .2 π$
$= \frac{10^{- 7} \times 4 π \times 2 π}{4 π \times 3 \times 10^{- 7}} (\frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}})$
$B = 125 μ T$

AP EAMCET-07.07.2022

Moving Charges & Magnetism

153118 A bar magnet with a magnetic moment $5.0 {Am}^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T$ . The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

1 J

3

2 J

4 Zero

Explanation:

A Given,
$Magnetic moment (M) = 5 {Am}^{2}$
$Magnetic field (B) = 0.4 T$
We know that,
$W = \int τ . d θ$
$W = \int_{0}^{π} MB \sin θ d θ$
$W = MB \times [- \cos]_{0}^{π}$
$W = - 0.4 \times 5 [\cos 180^{\circ} - \cos 0^{\circ}]$
$W = - 0.4 \times 5 \times (- 2)$
$W = 4 J$

JEE Main-31.01.2023

Moving Charges & Magnetism

153119 A uniform current is flowing along the length of an infinite, straight, thin, hollow cylinder of radius $R$ . The magnetic field $B$ produced at a perpendicular distance $d$ from the axis of the cylinder is plotted in a graph. Which of the following figures looks like the plot?

1

2

3

4

Explanation:

C Let radius of hollow cylinder is $R$

For $d < R, B \times 2 π d = μ_{0} I_{in}$
$(∵$ Magnetic field inside conductor is zero)
$B = 0$
$(I_{in} = 0)$
For $d > R B \times 2 π d = μ_{0} I$
$B = \frac{μ_{0} I}{2 π d} \Rightarrow B \propto \frac{1}{d}$

JEE Main-28.06.2022

Moving Charges & Magnetism

153121 A short bar magnet produces a magnetic field of $6.4 \times 10^{- 5} T$ at a distance of $20 cm$ from the center of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 cm$ from the center of the magnet on the axis, is

1

4.8 \times 10^{- 5} T

2

3.2 \times 10^{- 5} T

3

1.6 \times 10^{- 5} T

4

6.4 \times 10^{- 5} T

Explanation:

C Given,
Normal bisector distance $(d_{1}) = 20 cm = 0.2 m$
Axial distance $(d_{2}) = 40 cm = 0.4 m$
Magnetic field at its bisector
$B_{bisector} = \frac{μ_{0}}{4 π} \times \frac{M}{d_{1}^{3}}$
Magnetic field at its axis
$B_{axis} = \frac{μ_{0}}{4 π} \times \frac{2 M}{d_{2}^{3}}$
Ratio of magnetic fields,
$\frac{B_{axis}}{B_{bisector}} = \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = B_{bisector} \times \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = 6.4 \times 10^{- 5} \times \frac{2 \times (0.2)^{3}}{(0.4)^{3}}$
$B_{axis} = 1.6 \times 10^{- 5} T$

AP EAMCET-07.07.2022

Moving Charges & Magnetism

153122 The magnetic field due to current carrying circular loop of radius $6 cm$ at a point on the axis at a distance of $8 cm$ from centre is $27 μ T$ . The magnetic field at the centre of the current carrying loop is.

1

75 μ T

.

2

125 μ T

3

150 μ T

.

4

250 μ T

.

Explanation:

B Given,
Circular loop of radius $(r) = 6 cm = 6 \times 10^{- 2} m$
Axial distance $(x) = 8 cm = 8 \times 10^{- 2} m$
Magnetic field at centre $= 27 μ T = 27 \times 10^{- 6} T$
$B = \frac{μ_{o}}{4 π} I 2 π \frac{r^{2}}{{(r^{2} + x^{2})}^{3 / 2}}$
$27 \times 10^{- 6} = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{I \times 2 π \times {(6 \times 10^{- 2})}^{2}}{{({(6 \times 10^{- 2})}^{2} + {(8 \times 10^{- 2})}^{2})}^{3 / 2}}$
$I = \frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}}$
Magnetic field at the centre
$B = \frac{μ_{0} I}{4 π r} .2 π$
$= \frac{10^{- 7} \times 4 π \times 2 π}{4 π \times 3 \times 10^{- 7}} (\frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}})$
$B = 125 μ T$

AP EAMCET-07.07.2022

Moving Charges & Magnetism

153118 A bar magnet with a magnetic moment $5.0 {Am}^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T$ . The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

1 J

3

2 J

4 Zero

Explanation:

A Given,
$Magnetic moment (M) = 5 {Am}^{2}$
$Magnetic field (B) = 0.4 T$
We know that,
$W = \int τ . d θ$
$W = \int_{0}^{π} MB \sin θ d θ$
$W = MB \times [- \cos]_{0}^{π}$
$W = - 0.4 \times 5 [\cos 180^{\circ} - \cos 0^{\circ}]$
$W = - 0.4 \times 5 \times (- 2)$
$W = 4 J$

JEE Main-31.01.2023

Moving Charges & Magnetism

153119 A uniform current is flowing along the length of an infinite, straight, thin, hollow cylinder of radius $R$ . The magnetic field $B$ produced at a perpendicular distance $d$ from the axis of the cylinder is plotted in a graph. Which of the following figures looks like the plot?

1

2

3

4

Explanation:

C Let radius of hollow cylinder is $R$

For $d < R, B \times 2 π d = μ_{0} I_{in}$
$(∵$ Magnetic field inside conductor is zero)
$B = 0$
$(I_{in} = 0)$
For $d > R B \times 2 π d = μ_{0} I$
$B = \frac{μ_{0} I}{2 π d} \Rightarrow B \propto \frac{1}{d}$

JEE Main-28.06.2022

Moving Charges & Magnetism

153121 A short bar magnet produces a magnetic field of $6.4 \times 10^{- 5} T$ at a distance of $20 cm$ from the center of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 cm$ from the center of the magnet on the axis, is

1

4.8 \times 10^{- 5} T

2

3.2 \times 10^{- 5} T

3

1.6 \times 10^{- 5} T

4

6.4 \times 10^{- 5} T

Explanation:

C Given,
Normal bisector distance $(d_{1}) = 20 cm = 0.2 m$
Axial distance $(d_{2}) = 40 cm = 0.4 m$
Magnetic field at its bisector
$B_{bisector} = \frac{μ_{0}}{4 π} \times \frac{M}{d_{1}^{3}}$
Magnetic field at its axis
$B_{axis} = \frac{μ_{0}}{4 π} \times \frac{2 M}{d_{2}^{3}}$
Ratio of magnetic fields,
$\frac{B_{axis}}{B_{bisector}} = \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = B_{bisector} \times \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = 6.4 \times 10^{- 5} \times \frac{2 \times (0.2)^{3}}{(0.4)^{3}}$
$B_{axis} = 1.6 \times 10^{- 5} T$

AP EAMCET-07.07.2022

Moving Charges & Magnetism

153122 The magnetic field due to current carrying circular loop of radius $6 cm$ at a point on the axis at a distance of $8 cm$ from centre is $27 μ T$ . The magnetic field at the centre of the current carrying loop is.

1

75 μ T

.

2

125 μ T

3

150 μ T

.

4

250 μ T

.

Explanation:

B Given,
Circular loop of radius $(r) = 6 cm = 6 \times 10^{- 2} m$
Axial distance $(x) = 8 cm = 8 \times 10^{- 2} m$
Magnetic field at centre $= 27 μ T = 27 \times 10^{- 6} T$
$B = \frac{μ_{o}}{4 π} I 2 π \frac{r^{2}}{{(r^{2} + x^{2})}^{3 / 2}}$
$27 \times 10^{- 6} = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{I \times 2 π \times {(6 \times 10^{- 2})}^{2}}{{({(6 \times 10^{- 2})}^{2} + {(8 \times 10^{- 2})}^{2})}^{3 / 2}}$
$I = \frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}}$
Magnetic field at the centre
$B = \frac{μ_{0} I}{4 π r} .2 π$
$= \frac{10^{- 7} \times 4 π \times 2 π}{4 π \times 3 \times 10^{- 7}} (\frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}})$
$B = 125 μ T$

AP EAMCET-07.07.2022

Moving Charges & Magnetism

153118 A bar magnet with a magnetic moment $5.0 {Am}^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T$ . The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

1 J

3

2 J

4 Zero

Explanation:

A Given,
$Magnetic moment (M) = 5 {Am}^{2}$
$Magnetic field (B) = 0.4 T$
We know that,
$W = \int τ . d θ$
$W = \int_{0}^{π} MB \sin θ d θ$
$W = MB \times [- \cos]_{0}^{π}$
$W = - 0.4 \times 5 [\cos 180^{\circ} - \cos 0^{\circ}]$
$W = - 0.4 \times 5 \times (- 2)$
$W = 4 J$

JEE Main-31.01.2023

Moving Charges & Magnetism

153119 A uniform current is flowing along the length of an infinite, straight, thin, hollow cylinder of radius $R$ . The magnetic field $B$ produced at a perpendicular distance $d$ from the axis of the cylinder is plotted in a graph. Which of the following figures looks like the plot?

1

2

3

4

Explanation:

C Let radius of hollow cylinder is $R$

For $d < R, B \times 2 π d = μ_{0} I_{in}$
$(∵$ Magnetic field inside conductor is zero)
$B = 0$
$(I_{in} = 0)$
For $d > R B \times 2 π d = μ_{0} I$
$B = \frac{μ_{0} I}{2 π d} \Rightarrow B \propto \frac{1}{d}$

JEE Main-28.06.2022

Moving Charges & Magnetism

153121 A short bar magnet produces a magnetic field of $6.4 \times 10^{- 5} T$ at a distance of $20 cm$ from the center of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 cm$ from the center of the magnet on the axis, is

1

4.8 \times 10^{- 5} T

2

3.2 \times 10^{- 5} T

3

1.6 \times 10^{- 5} T

4

6.4 \times 10^{- 5} T

Explanation:

C Given,
Normal bisector distance $(d_{1}) = 20 cm = 0.2 m$
Axial distance $(d_{2}) = 40 cm = 0.4 m$
Magnetic field at its bisector
$B_{bisector} = \frac{μ_{0}}{4 π} \times \frac{M}{d_{1}^{3}}$
Magnetic field at its axis
$B_{axis} = \frac{μ_{0}}{4 π} \times \frac{2 M}{d_{2}^{3}}$
Ratio of magnetic fields,
$\frac{B_{axis}}{B_{bisector}} = \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = B_{bisector} \times \frac{2 d_{1}^{3}}{d_{2}^{3}}$
$B_{axis} = 6.4 \times 10^{- 5} \times \frac{2 \times (0.2)^{3}}{(0.4)^{3}}$
$B_{axis} = 1.6 \times 10^{- 5} T$

AP EAMCET-07.07.2022

Moving Charges & Magnetism

153122 The magnetic field due to current carrying circular loop of radius $6 cm$ at a point on the axis at a distance of $8 cm$ from centre is $27 μ T$ . The magnetic field at the centre of the current carrying loop is.

1

75 μ T

.

2

125 μ T

3

150 μ T

.

4

250 μ T

.

Explanation:

B Given,
Circular loop of radius $(r) = 6 cm = 6 \times 10^{- 2} m$
Axial distance $(x) = 8 cm = 8 \times 10^{- 2} m$
Magnetic field at centre $= 27 μ T = 27 \times 10^{- 6} T$
$B = \frac{μ_{o}}{4 π} I 2 π \frac{r^{2}}{{(r^{2} + x^{2})}^{3 / 2}}$
$27 \times 10^{- 6} = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{I \times 2 π \times {(6 \times 10^{- 2})}^{2}}{{({(6 \times 10^{- 2})}^{2} + {(8 \times 10^{- 2})}^{2})}^{3 / 2}}$
$I = \frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}}$
Magnetic field at the centre
$B = \frac{μ_{0} I}{4 π r} .2 π$
$= \frac{10^{- 7} \times 4 π \times 2 π}{4 π \times 3 \times 10^{- 7}} (\frac{27 \times 10^{- 6} \times {(100 \times 10^{- 4})}^{3 / 2}}{2 π \times {(6 \times 10^{- 2})}^{2}})$
$B = 125 μ T$

AP EAMCET-07.07.2022

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