QBManager

Moving Charges & Magnetism

153222 Two short bar magnets each of magnetic moment of $9 {Am}^{2}$ are placed such that one is at $x = - 3 cm$ and the other at $y = - 3 cm$ . If their magnetic moments are directed along positive and negative $X$ -directions respectively, then the resultant magnetic field at the origin is

1

100 T

2

10 T

3

0.1 T

4

0.001 T

Explanation:

C Given, magnetic moment $(M) = 9 {Am}^{2}$

At the origin magnetic field direction will be same
$B = B_{1} + B_{2}$
Magnetic field due to $m_{1}$ (axial point)-
$B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 M}{r^{3}} (∵ \frac{μ_{0}}{4 π} = 10^{- 7})$
$B_{1} = \frac{10^{- 7} \times 2 \times 9}{27 \times 10^{- 6}}$
$B_{1} = 10^{- 7} \times 10^{6} \times \frac{2}{3}$
$B_{1} = \frac{2}{3} \times 10^{- 1}$
Magnetic field due to $m_{2}$ (equatorial point)-
$B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{M}{r^{3}}$
$= \frac{10^{- 7} \times 9}{{(3 \times 10^{- 2})}^{3}} = \frac{10^{- 7} \times 9}{27 \times 10^{- 6}}$
$B_{2} = \frac{1}{3} \times 10^{- 7} \times 10^{6} = \frac{1}{3} \times 10^{- 1}$
Putting the value of $B_{1} & B_{2}$ in equation (i), we get-
$B = B_{1} + B_{2}$
$= \frac{2}{3} 10^{- 1} + \frac{1}{3} 10^{- 1}$
$B = \frac{1}{10} = 0.1 T$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153223 Two infinitely long wires carry currents $4 A$ and $3 A$ placed along $X$ -axis and $Y$ -axis respectively. Magnetic field at a point $P (0, 0, d)$ m will be ...... T.

1

\frac{4 μ_{0}}{2 π d}

2

\frac{3 μ_{0}}{2 π d}

3

\frac{7 μ_{0}}{2 π d}

4

\frac{5 μ_{0}}{2 π d}

Explanation:

D Given, $I_{1} = 4 A, I_{2} = 3 A$

Resultant $B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$∵ B_{1} = (\frac{μ_{0} I_{1}}{2 π d})$ and $B_{2} = (\frac{μ_{0} I_{2}}{2 π d})$
Putting the value of $B_{1} & B_{2}$ in equation (i)
$B = \sqrt{{(\frac{μ_{0} I_{1}}{2 π d})}^{2} + {(\frac{μ_{0} I_{2}}{2 π d})}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{I_{1}^{2} + I_{2}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{(4)^{2} + 3^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{25}$
$B = \frac{μ_{0}}{2 π d} \times 5 = \frac{5 μ_{o}}{2 π d}$
$B = \frac{5 μ_{0}}{2 π d}$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153226 A bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along positive $X$ axis. Likewise, second bar magnet of magnetic moment $M$ is placed at a distance $2 D$ on positive $Y$ -axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin is $| B | = α [\frac{μ_{0}}{4 π} \frac{M}{D^{3}}]$ .
The value of $α$ must be (assume $D >> l$ , where $l$ is the length of magnets)

1 2

2

\frac{15}{8}

3

\frac{17}{8}

4

\frac{9}{8}

Explanation:

B Magnetic field at axial point (D $>> l$ )
$B_{1} = \frac{2 μ_{0} M}{4 π D^{3}} (From S to N)$
Magnetic field at equatorial line $(2 D >> l)$
$B_{2} = \frac{μ_{0} M}{4 π (2 D)^{3}} (From N to S)$
$B_{2} = \frac{1}{8} \frac{μ_{0} M}{4 π D^{3}}$
So, resultant magnetic field
$B = B_{1} - B_{2} (From S to N)$
$B = \frac{μ_{0} M}{4 π D^{3}} (2 - \frac{1}{8})$
$\frac{15}{8} \frac{μ_{0} M}{4 π D^{3}}$ (In magnitude)

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153227 A steady current I flows through a wire with one end at $O$ and the other end extending upto infinity as shown in the figure. The magnetic field at a point $P$ , located at a distance $d$ from $O$ is

1

\frac{μ_{0} I}{4 π d \cos α} (1 - \sin α)

2

\frac{μ_{0} I}{2 π d \cos α} (1 - \sin α)

3

\frac{μ_{0} I}{4 π d}

4

\frac{μ_{0} I}{4 π d \sin α} (1 - \cos α)

Explanation:

D
From result of field a long wire we have
$B = \frac{μ_{0}}{4 π} \times \frac{I}{r} \times (\sin θ_{1} - \sin θ_{2})$
$B = \frac{μ_{0} I}{4 π r} [\sin (90^{\circ} - α) + \sin 90^{\circ}]$
$B = \frac{μ_{0} I}{4 π r} [- \cos α + 1]$
$So, B = \frac{μ_{0} I}{4 π d \sin α} (1 - \cos α) (∴ r = d \sin α)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153228 Consider a current carrying wire shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is

1

\frac{μ_{0}}{4 π} \frac{i}{R} (2 + π)

2

\frac{μ_{0} i}{2 π R}

3

\frac{μ_{0}}{2} \frac{i}{R}

4

\frac{μ_{0}}{4} \frac{i}{R}

Explanation:

A Magnetic field at point $O$
$B = \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} + \frac{μ_{0} i}{4 π R}$
$B = \frac{μ_{0} i}{4 π R} (2 + π)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153222 Two short bar magnets each of magnetic moment of $9 {Am}^{2}$ are placed such that one is at $x = - 3 cm$ and the other at $y = - 3 cm$ . If their magnetic moments are directed along positive and negative $X$ -directions respectively, then the resultant magnetic field at the origin is

1

100 T

2

10 T

3

0.1 T

4

0.001 T

Explanation:

C Given, magnetic moment $(M) = 9 {Am}^{2}$

At the origin magnetic field direction will be same
$B = B_{1} + B_{2}$
Magnetic field due to $m_{1}$ (axial point)-
$B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 M}{r^{3}} (∵ \frac{μ_{0}}{4 π} = 10^{- 7})$
$B_{1} = \frac{10^{- 7} \times 2 \times 9}{27 \times 10^{- 6}}$
$B_{1} = 10^{- 7} \times 10^{6} \times \frac{2}{3}$
$B_{1} = \frac{2}{3} \times 10^{- 1}$
Magnetic field due to $m_{2}$ (equatorial point)-
$B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{M}{r^{3}}$
$= \frac{10^{- 7} \times 9}{{(3 \times 10^{- 2})}^{3}} = \frac{10^{- 7} \times 9}{27 \times 10^{- 6}}$
$B_{2} = \frac{1}{3} \times 10^{- 7} \times 10^{6} = \frac{1}{3} \times 10^{- 1}$
Putting the value of $B_{1} & B_{2}$ in equation (i), we get-
$B = B_{1} + B_{2}$
$= \frac{2}{3} 10^{- 1} + \frac{1}{3} 10^{- 1}$
$B = \frac{1}{10} = 0.1 T$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153223 Two infinitely long wires carry currents $4 A$ and $3 A$ placed along $X$ -axis and $Y$ -axis respectively. Magnetic field at a point $P (0, 0, d)$ m will be ...... T.

1

\frac{4 μ_{0}}{2 π d}

2

\frac{3 μ_{0}}{2 π d}

3

\frac{7 μ_{0}}{2 π d}

4

\frac{5 μ_{0}}{2 π d}

Explanation:

D Given, $I_{1} = 4 A, I_{2} = 3 A$

Resultant $B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$∵ B_{1} = (\frac{μ_{0} I_{1}}{2 π d})$ and $B_{2} = (\frac{μ_{0} I_{2}}{2 π d})$
Putting the value of $B_{1} & B_{2}$ in equation (i)
$B = \sqrt{{(\frac{μ_{0} I_{1}}{2 π d})}^{2} + {(\frac{μ_{0} I_{2}}{2 π d})}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{I_{1}^{2} + I_{2}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{(4)^{2} + 3^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{25}$
$B = \frac{μ_{0}}{2 π d} \times 5 = \frac{5 μ_{o}}{2 π d}$
$B = \frac{5 μ_{0}}{2 π d}$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153226 A bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along positive $X$ axis. Likewise, second bar magnet of magnetic moment $M$ is placed at a distance $2 D$ on positive $Y$ -axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin is $| B | = α [\frac{μ_{0}}{4 π} \frac{M}{D^{3}}]$ .
The value of $α$ must be (assume $D >> l$ , where $l$ is the length of magnets)

1 2

2

\frac{15}{8}

3

\frac{17}{8}

4

\frac{9}{8}

Explanation:

B Magnetic field at axial point (D $>> l$ )
$B_{1} = \frac{2 μ_{0} M}{4 π D^{3}} (From S to N)$
Magnetic field at equatorial line $(2 D >> l)$
$B_{2} = \frac{μ_{0} M}{4 π (2 D)^{3}} (From N to S)$
$B_{2} = \frac{1}{8} \frac{μ_{0} M}{4 π D^{3}}$
So, resultant magnetic field
$B = B_{1} - B_{2} (From S to N)$
$B = \frac{μ_{0} M}{4 π D^{3}} (2 - \frac{1}{8})$
$\frac{15}{8} \frac{μ_{0} M}{4 π D^{3}}$ (In magnitude)

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153227 A steady current I flows through a wire with one end at $O$ and the other end extending upto infinity as shown in the figure. The magnetic field at a point $P$ , located at a distance $d$ from $O$ is

1

\frac{μ_{0} I}{4 π d \cos α} (1 - \sin α)

2

\frac{μ_{0} I}{2 π d \cos α} (1 - \sin α)

3

\frac{μ_{0} I}{4 π d}

4

\frac{μ_{0} I}{4 π d \sin α} (1 - \cos α)

Explanation:

D
From result of field a long wire we have
$B = \frac{μ_{0}}{4 π} \times \frac{I}{r} \times (\sin θ_{1} - \sin θ_{2})$
$B = \frac{μ_{0} I}{4 π r} [\sin (90^{\circ} - α) + \sin 90^{\circ}]$
$B = \frac{μ_{0} I}{4 π r} [- \cos α + 1]$
$So, B = \frac{μ_{0} I}{4 π d \sin α} (1 - \cos α) (∴ r = d \sin α)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153228 Consider a current carrying wire shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is

1

\frac{μ_{0}}{4 π} \frac{i}{R} (2 + π)

2

\frac{μ_{0} i}{2 π R}

3

\frac{μ_{0}}{2} \frac{i}{R}

4

\frac{μ_{0}}{4} \frac{i}{R}

Explanation:

A Magnetic field at point $O$
$B = \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} + \frac{μ_{0} i}{4 π R}$
$B = \frac{μ_{0} i}{4 π R} (2 + π)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153222 Two short bar magnets each of magnetic moment of $9 {Am}^{2}$ are placed such that one is at $x = - 3 cm$ and the other at $y = - 3 cm$ . If their magnetic moments are directed along positive and negative $X$ -directions respectively, then the resultant magnetic field at the origin is

1

100 T

2

10 T

3

0.1 T

4

0.001 T

Explanation:

C Given, magnetic moment $(M) = 9 {Am}^{2}$

At the origin magnetic field direction will be same
$B = B_{1} + B_{2}$
Magnetic field due to $m_{1}$ (axial point)-
$B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 M}{r^{3}} (∵ \frac{μ_{0}}{4 π} = 10^{- 7})$
$B_{1} = \frac{10^{- 7} \times 2 \times 9}{27 \times 10^{- 6}}$
$B_{1} = 10^{- 7} \times 10^{6} \times \frac{2}{3}$
$B_{1} = \frac{2}{3} \times 10^{- 1}$
Magnetic field due to $m_{2}$ (equatorial point)-
$B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{M}{r^{3}}$
$= \frac{10^{- 7} \times 9}{{(3 \times 10^{- 2})}^{3}} = \frac{10^{- 7} \times 9}{27 \times 10^{- 6}}$
$B_{2} = \frac{1}{3} \times 10^{- 7} \times 10^{6} = \frac{1}{3} \times 10^{- 1}$
Putting the value of $B_{1} & B_{2}$ in equation (i), we get-
$B = B_{1} + B_{2}$
$= \frac{2}{3} 10^{- 1} + \frac{1}{3} 10^{- 1}$
$B = \frac{1}{10} = 0.1 T$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153223 Two infinitely long wires carry currents $4 A$ and $3 A$ placed along $X$ -axis and $Y$ -axis respectively. Magnetic field at a point $P (0, 0, d)$ m will be ...... T.

1

\frac{4 μ_{0}}{2 π d}

2

\frac{3 μ_{0}}{2 π d}

3

\frac{7 μ_{0}}{2 π d}

4

\frac{5 μ_{0}}{2 π d}

Explanation:

D Given, $I_{1} = 4 A, I_{2} = 3 A$

Resultant $B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$∵ B_{1} = (\frac{μ_{0} I_{1}}{2 π d})$ and $B_{2} = (\frac{μ_{0} I_{2}}{2 π d})$
Putting the value of $B_{1} & B_{2}$ in equation (i)
$B = \sqrt{{(\frac{μ_{0} I_{1}}{2 π d})}^{2} + {(\frac{μ_{0} I_{2}}{2 π d})}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{I_{1}^{2} + I_{2}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{(4)^{2} + 3^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{25}$
$B = \frac{μ_{0}}{2 π d} \times 5 = \frac{5 μ_{o}}{2 π d}$
$B = \frac{5 μ_{0}}{2 π d}$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153226 A bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along positive $X$ axis. Likewise, second bar magnet of magnetic moment $M$ is placed at a distance $2 D$ on positive $Y$ -axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin is $| B | = α [\frac{μ_{0}}{4 π} \frac{M}{D^{3}}]$ .
The value of $α$ must be (assume $D >> l$ , where $l$ is the length of magnets)

1 2

2

\frac{15}{8}

3

\frac{17}{8}

4

\frac{9}{8}

Explanation:

B Magnetic field at axial point (D $>> l$ )
$B_{1} = \frac{2 μ_{0} M}{4 π D^{3}} (From S to N)$
Magnetic field at equatorial line $(2 D >> l)$
$B_{2} = \frac{μ_{0} M}{4 π (2 D)^{3}} (From N to S)$
$B_{2} = \frac{1}{8} \frac{μ_{0} M}{4 π D^{3}}$
So, resultant magnetic field
$B = B_{1} - B_{2} (From S to N)$
$B = \frac{μ_{0} M}{4 π D^{3}} (2 - \frac{1}{8})$
$\frac{15}{8} \frac{μ_{0} M}{4 π D^{3}}$ (In magnitude)

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153227 A steady current I flows through a wire with one end at $O$ and the other end extending upto infinity as shown in the figure. The magnetic field at a point $P$ , located at a distance $d$ from $O$ is

1

\frac{μ_{0} I}{4 π d \cos α} (1 - \sin α)

2

\frac{μ_{0} I}{2 π d \cos α} (1 - \sin α)

3

\frac{μ_{0} I}{4 π d}

4

\frac{μ_{0} I}{4 π d \sin α} (1 - \cos α)

Explanation:

D
From result of field a long wire we have
$B = \frac{μ_{0}}{4 π} \times \frac{I}{r} \times (\sin θ_{1} - \sin θ_{2})$
$B = \frac{μ_{0} I}{4 π r} [\sin (90^{\circ} - α) + \sin 90^{\circ}]$
$B = \frac{μ_{0} I}{4 π r} [- \cos α + 1]$
$So, B = \frac{μ_{0} I}{4 π d \sin α} (1 - \cos α) (∴ r = d \sin α)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153228 Consider a current carrying wire shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is

1

\frac{μ_{0}}{4 π} \frac{i}{R} (2 + π)

2

\frac{μ_{0} i}{2 π R}

3

\frac{μ_{0}}{2} \frac{i}{R}

4

\frac{μ_{0}}{4} \frac{i}{R}

Explanation:

A Magnetic field at point $O$
$B = \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} + \frac{μ_{0} i}{4 π R}$
$B = \frac{μ_{0} i}{4 π R} (2 + π)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153222 Two short bar magnets each of magnetic moment of $9 {Am}^{2}$ are placed such that one is at $x = - 3 cm$ and the other at $y = - 3 cm$ . If their magnetic moments are directed along positive and negative $X$ -directions respectively, then the resultant magnetic field at the origin is

1

100 T

2

10 T

3

0.1 T

4

0.001 T

Explanation:

C Given, magnetic moment $(M) = 9 {Am}^{2}$

At the origin magnetic field direction will be same
$B = B_{1} + B_{2}$
Magnetic field due to $m_{1}$ (axial point)-
$B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 M}{r^{3}} (∵ \frac{μ_{0}}{4 π} = 10^{- 7})$
$B_{1} = \frac{10^{- 7} \times 2 \times 9}{27 \times 10^{- 6}}$
$B_{1} = 10^{- 7} \times 10^{6} \times \frac{2}{3}$
$B_{1} = \frac{2}{3} \times 10^{- 1}$
Magnetic field due to $m_{2}$ (equatorial point)-
$B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{M}{r^{3}}$
$= \frac{10^{- 7} \times 9}{{(3 \times 10^{- 2})}^{3}} = \frac{10^{- 7} \times 9}{27 \times 10^{- 6}}$
$B_{2} = \frac{1}{3} \times 10^{- 7} \times 10^{6} = \frac{1}{3} \times 10^{- 1}$
Putting the value of $B_{1} & B_{2}$ in equation (i), we get-
$B = B_{1} + B_{2}$
$= \frac{2}{3} 10^{- 1} + \frac{1}{3} 10^{- 1}$
$B = \frac{1}{10} = 0.1 T$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153223 Two infinitely long wires carry currents $4 A$ and $3 A$ placed along $X$ -axis and $Y$ -axis respectively. Magnetic field at a point $P (0, 0, d)$ m will be ...... T.

1

\frac{4 μ_{0}}{2 π d}

2

\frac{3 μ_{0}}{2 π d}

3

\frac{7 μ_{0}}{2 π d}

4

\frac{5 μ_{0}}{2 π d}

Explanation:

D Given, $I_{1} = 4 A, I_{2} = 3 A$

Resultant $B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$∵ B_{1} = (\frac{μ_{0} I_{1}}{2 π d})$ and $B_{2} = (\frac{μ_{0} I_{2}}{2 π d})$
Putting the value of $B_{1} & B_{2}$ in equation (i)
$B = \sqrt{{(\frac{μ_{0} I_{1}}{2 π d})}^{2} + {(\frac{μ_{0} I_{2}}{2 π d})}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{I_{1}^{2} + I_{2}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{(4)^{2} + 3^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{25}$
$B = \frac{μ_{0}}{2 π d} \times 5 = \frac{5 μ_{o}}{2 π d}$
$B = \frac{5 μ_{0}}{2 π d}$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153226 A bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along positive $X$ axis. Likewise, second bar magnet of magnetic moment $M$ is placed at a distance $2 D$ on positive $Y$ -axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin is $| B | = α [\frac{μ_{0}}{4 π} \frac{M}{D^{3}}]$ .
The value of $α$ must be (assume $D >> l$ , where $l$ is the length of magnets)

1 2

2

\frac{15}{8}

3

\frac{17}{8}

4

\frac{9}{8}

Explanation:

B Magnetic field at axial point (D $>> l$ )
$B_{1} = \frac{2 μ_{0} M}{4 π D^{3}} (From S to N)$
Magnetic field at equatorial line $(2 D >> l)$
$B_{2} = \frac{μ_{0} M}{4 π (2 D)^{3}} (From N to S)$
$B_{2} = \frac{1}{8} \frac{μ_{0} M}{4 π D^{3}}$
So, resultant magnetic field
$B = B_{1} - B_{2} (From S to N)$
$B = \frac{μ_{0} M}{4 π D^{3}} (2 - \frac{1}{8})$
$\frac{15}{8} \frac{μ_{0} M}{4 π D^{3}}$ (In magnitude)

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153227 A steady current I flows through a wire with one end at $O$ and the other end extending upto infinity as shown in the figure. The magnetic field at a point $P$ , located at a distance $d$ from $O$ is

1

\frac{μ_{0} I}{4 π d \cos α} (1 - \sin α)

2

\frac{μ_{0} I}{2 π d \cos α} (1 - \sin α)

3

\frac{μ_{0} I}{4 π d}

4

\frac{μ_{0} I}{4 π d \sin α} (1 - \cos α)

Explanation:

D
From result of field a long wire we have
$B = \frac{μ_{0}}{4 π} \times \frac{I}{r} \times (\sin θ_{1} - \sin θ_{2})$
$B = \frac{μ_{0} I}{4 π r} [\sin (90^{\circ} - α) + \sin 90^{\circ}]$
$B = \frac{μ_{0} I}{4 π r} [- \cos α + 1]$
$So, B = \frac{μ_{0} I}{4 π d \sin α} (1 - \cos α) (∴ r = d \sin α)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153228 Consider a current carrying wire shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is

1

\frac{μ_{0}}{4 π} \frac{i}{R} (2 + π)

2

\frac{μ_{0} i}{2 π R}

3

\frac{μ_{0}}{2} \frac{i}{R}

4

\frac{μ_{0}}{4} \frac{i}{R}

Explanation:

A Magnetic field at point $O$
$B = \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} + \frac{μ_{0} i}{4 π R}$
$B = \frac{μ_{0} i}{4 π R} (2 + π)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153222 Two short bar magnets each of magnetic moment of $9 {Am}^{2}$ are placed such that one is at $x = - 3 cm$ and the other at $y = - 3 cm$ . If their magnetic moments are directed along positive and negative $X$ -directions respectively, then the resultant magnetic field at the origin is

1

100 T

2

10 T

3

0.1 T

4

0.001 T

Explanation:

C Given, magnetic moment $(M) = 9 {Am}^{2}$

At the origin magnetic field direction will be same
$B = B_{1} + B_{2}$
Magnetic field due to $m_{1}$ (axial point)-
$B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 M}{r^{3}} (∵ \frac{μ_{0}}{4 π} = 10^{- 7})$
$B_{1} = \frac{10^{- 7} \times 2 \times 9}{27 \times 10^{- 6}}$
$B_{1} = 10^{- 7} \times 10^{6} \times \frac{2}{3}$
$B_{1} = \frac{2}{3} \times 10^{- 1}$
Magnetic field due to $m_{2}$ (equatorial point)-
$B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{M}{r^{3}}$
$= \frac{10^{- 7} \times 9}{{(3 \times 10^{- 2})}^{3}} = \frac{10^{- 7} \times 9}{27 \times 10^{- 6}}$
$B_{2} = \frac{1}{3} \times 10^{- 7} \times 10^{6} = \frac{1}{3} \times 10^{- 1}$
Putting the value of $B_{1} & B_{2}$ in equation (i), we get-
$B = B_{1} + B_{2}$
$= \frac{2}{3} 10^{- 1} + \frac{1}{3} 10^{- 1}$
$B = \frac{1}{10} = 0.1 T$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153223 Two infinitely long wires carry currents $4 A$ and $3 A$ placed along $X$ -axis and $Y$ -axis respectively. Magnetic field at a point $P (0, 0, d)$ m will be ...... T.

1

\frac{4 μ_{0}}{2 π d}

2

\frac{3 μ_{0}}{2 π d}

3

\frac{7 μ_{0}}{2 π d}

4

\frac{5 μ_{0}}{2 π d}

Explanation:

D Given, $I_{1} = 4 A, I_{2} = 3 A$

Resultant $B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$∵ B_{1} = (\frac{μ_{0} I_{1}}{2 π d})$ and $B_{2} = (\frac{μ_{0} I_{2}}{2 π d})$
Putting the value of $B_{1} & B_{2}$ in equation (i)
$B = \sqrt{{(\frac{μ_{0} I_{1}}{2 π d})}^{2} + {(\frac{μ_{0} I_{2}}{2 π d})}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{I_{1}^{2} + I_{2}^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{(4)^{2} + 3^{2}}$
$B = \frac{μ_{0}}{2 π d} \sqrt{25}$
$B = \frac{μ_{0}}{2 π d} \times 5 = \frac{5 μ_{o}}{2 π d}$
$B = \frac{5 μ_{0}}{2 π d}$

AP EAMCET (22.04.2018) Shift-1

Moving Charges & Magnetism

153226 A bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along positive $X$ axis. Likewise, second bar magnet of magnetic moment $M$ is placed at a distance $2 D$ on positive $Y$ -axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin is $| B | = α [\frac{μ_{0}}{4 π} \frac{M}{D^{3}}]$ .
The value of $α$ must be (assume $D >> l$ , where $l$ is the length of magnets)

1 2

2

\frac{15}{8}

3

\frac{17}{8}

4

\frac{9}{8}

Explanation:

B Magnetic field at axial point (D $>> l$ )
$B_{1} = \frac{2 μ_{0} M}{4 π D^{3}} (From S to N)$
Magnetic field at equatorial line $(2 D >> l)$
$B_{2} = \frac{μ_{0} M}{4 π (2 D)^{3}} (From N to S)$
$B_{2} = \frac{1}{8} \frac{μ_{0} M}{4 π D^{3}}$
So, resultant magnetic field
$B = B_{1} - B_{2} (From S to N)$
$B = \frac{μ_{0} M}{4 π D^{3}} (2 - \frac{1}{8})$
$\frac{15}{8} \frac{μ_{0} M}{4 π D^{3}}$ (In magnitude)

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153227 A steady current I flows through a wire with one end at $O$ and the other end extending upto infinity as shown in the figure. The magnetic field at a point $P$ , located at a distance $d$ from $O$ is

1

\frac{μ_{0} I}{4 π d \cos α} (1 - \sin α)

2

\frac{μ_{0} I}{2 π d \cos α} (1 - \sin α)

3

\frac{μ_{0} I}{4 π d}

4

\frac{μ_{0} I}{4 π d \sin α} (1 - \cos α)

Explanation:

D
From result of field a long wire we have
$B = \frac{μ_{0}}{4 π} \times \frac{I}{r} \times (\sin θ_{1} - \sin θ_{2})$
$B = \frac{μ_{0} I}{4 π r} [\sin (90^{\circ} - α) + \sin 90^{\circ}]$
$B = \frac{μ_{0} I}{4 π r} [- \cos α + 1]$
$So, B = \frac{μ_{0} I}{4 π d \sin α} (1 - \cos α) (∴ r = d \sin α)$

TS- EAMCET-05.05.2018

Moving Charges & Magnetism

153228 Consider a current carrying wire shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is

1

\frac{μ_{0}}{4 π} \frac{i}{R} (2 + π)

2

\frac{μ_{0} i}{2 π R}

3

\frac{μ_{0}}{2} \frac{i}{R}

4

\frac{μ_{0}}{4} \frac{i}{R}

Explanation:

A Magnetic field at point $O$
$B = \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} + \frac{μ_{0} i}{4 π R}$
$B = \frac{μ_{0} i}{4 π R} (2 + π)$

TS- EAMCET-05.05.2018

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