QBManager

Moving Charges & Magnetism

153179 A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 cm$ and $2 cm$ , respectively. When a current $1 A$ passes through the coil, the magnetic field at the centre of the coil is

1

2 π \ln (2) mT

2

\frac{π}{2} \ln (2) mT

3

π \ln (2) mT

4

\sqrt{2} π \ln (2) mT

Explanation:

A Given,
$N = 100$
$r_{1} = 1 cm = 1 \times 10^{- 2} m$
$r_{2} = 2 cm = 2 \times 10^{- 2} m$
$I = 1 A$
$B = \frac{μ_{0} NI}{2 (r_{2} - r_{1})} \ln (\frac{r_{2}}{r_{1}})$
$= \frac{4 π \times 10^{- 7} \times 100 \times 1}{2 \times (2 - 1) \times 10^{- 2}} \ln (\frac{2 \times 10^{- 2}}{1 \times 10^{- 2}})$
$B = 2 π \ln (2) \times 10^{- 3} T$
$B = 2 π \ln (2) mT$

TS- EAMCET-11.09.2020

Moving Charges & Magnetism

153180 The magnetic induction at point $O$ of the given infinitely long current carrying wire shown in the figure below is

1

\frac{μ_{0} I}{4 π R} (1 - \frac{3 π}{2})

2

\frac{μ_{0} I}{2 R (1 + π)}

3

\frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})

4

\frac{μ_{0} I}{4 π R}

Explanation:

C Magnetic field at point $O =$ Magnetic field at $O$ due to straight wire + Magnetic field of wire of loop
$B = \frac{μ_{0} I}{4 π R} + \frac{μ_{0} I}{4 π R} (\frac{3 π}{2})$
$B = \frac{μ_{0} I}{4 π R} + \frac{3}{8} \frac{μ_{0} I}{R}$
$B = \frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})$

Karnataka CET-2020

Moving Charges & Magnetism

153176 The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$ , is

1

\frac{μ_{0} I}{4 π r} T

2

\frac{μ_{o} I}{2 π r} T

3

\frac{μ_{o} I}{2 π R} T

4

\frac{μ_{o} I}{4 π R} T

Explanation:

B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $I$ by a conductor of Radius $R$ is given by
$B = \frac{μ_{0} I}{2 π r} T$

TS- EAMCET-09.09.2020

Moving Charges & Magnetism

153181 A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is

1 Zero

2

3 μ_{0} i / 32 R

, outward

3

3 μ_{0} i / 32 R

, inward

4

\frac{μ_{0} i}{2 R}

, inward

Explanation:

A Magnetic field due to $i_{1} = \frac{μ_{0} i_{1} θ_{1}}{2 R 2 π}$
$Magnetic field due to i_{2} = \frac{μ_{0} i_{2} θ_{2}}{2 R 2 π}$
For parallel combination-
$\frac{i_{1}}{i_{2}} = \frac{ρ l_{2}}{A} \times \frac{A}{ρ l_{1}} = \frac{l_{1}}{l_{2}}$
$\frac{i_{1}}{i_{2}} = \frac{\frac{1}{4} (2 π R)}{\frac{3}{4} (2 π R)}$
$i_{1} = \frac{i_{2}}{3}, i_{2} = 3 i_{1}$
Net magnetic field-
$= \frac{μ_{0} i_{1}}{2 R} (\frac{θ_{1}}{2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{θ_{2}}{2 π})$
$= \frac{μ_{0}}{2 R} (\frac{3 π}{2 \times 2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{π}{2 π}) \times \frac{1}{2}$
$= \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{i_{2}}{4}) = \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{3 i_{1}}{4}) = 0$

NEET Odisha-2019

Moving Charges & Magnetism

153179 A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 cm$ and $2 cm$ , respectively. When a current $1 A$ passes through the coil, the magnetic field at the centre of the coil is

1

2 π \ln (2) mT

2

\frac{π}{2} \ln (2) mT

3

π \ln (2) mT

4

\sqrt{2} π \ln (2) mT

Explanation:

A Given,
$N = 100$
$r_{1} = 1 cm = 1 \times 10^{- 2} m$
$r_{2} = 2 cm = 2 \times 10^{- 2} m$
$I = 1 A$
$B = \frac{μ_{0} NI}{2 (r_{2} - r_{1})} \ln (\frac{r_{2}}{r_{1}})$
$= \frac{4 π \times 10^{- 7} \times 100 \times 1}{2 \times (2 - 1) \times 10^{- 2}} \ln (\frac{2 \times 10^{- 2}}{1 \times 10^{- 2}})$
$B = 2 π \ln (2) \times 10^{- 3} T$
$B = 2 π \ln (2) mT$

TS- EAMCET-11.09.2020

Moving Charges & Magnetism

153180 The magnetic induction at point $O$ of the given infinitely long current carrying wire shown in the figure below is

1

\frac{μ_{0} I}{4 π R} (1 - \frac{3 π}{2})

2

\frac{μ_{0} I}{2 R (1 + π)}

3

\frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})

4

\frac{μ_{0} I}{4 π R}

Explanation:

C Magnetic field at point $O =$ Magnetic field at $O$ due to straight wire + Magnetic field of wire of loop
$B = \frac{μ_{0} I}{4 π R} + \frac{μ_{0} I}{4 π R} (\frac{3 π}{2})$
$B = \frac{μ_{0} I}{4 π R} + \frac{3}{8} \frac{μ_{0} I}{R}$
$B = \frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})$

Karnataka CET-2020

Moving Charges & Magnetism

153176 The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$ , is

1

\frac{μ_{0} I}{4 π r} T

2

\frac{μ_{o} I}{2 π r} T

3

\frac{μ_{o} I}{2 π R} T

4

\frac{μ_{o} I}{4 π R} T

Explanation:

B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $I$ by a conductor of Radius $R$ is given by
$B = \frac{μ_{0} I}{2 π r} T$

TS- EAMCET-09.09.2020

Moving Charges & Magnetism

153181 A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is

1 Zero

2

3 μ_{0} i / 32 R

, outward

3

3 μ_{0} i / 32 R

, inward

4

\frac{μ_{0} i}{2 R}

, inward

Explanation:

A Magnetic field due to $i_{1} = \frac{μ_{0} i_{1} θ_{1}}{2 R 2 π}$
$Magnetic field due to i_{2} = \frac{μ_{0} i_{2} θ_{2}}{2 R 2 π}$
For parallel combination-
$\frac{i_{1}}{i_{2}} = \frac{ρ l_{2}}{A} \times \frac{A}{ρ l_{1}} = \frac{l_{1}}{l_{2}}$
$\frac{i_{1}}{i_{2}} = \frac{\frac{1}{4} (2 π R)}{\frac{3}{4} (2 π R)}$
$i_{1} = \frac{i_{2}}{3}, i_{2} = 3 i_{1}$
Net magnetic field-
$= \frac{μ_{0} i_{1}}{2 R} (\frac{θ_{1}}{2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{θ_{2}}{2 π})$
$= \frac{μ_{0}}{2 R} (\frac{3 π}{2 \times 2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{π}{2 π}) \times \frac{1}{2}$
$= \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{i_{2}}{4}) = \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{3 i_{1}}{4}) = 0$

NEET Odisha-2019

Moving Charges & Magnetism

153179 A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 cm$ and $2 cm$ , respectively. When a current $1 A$ passes through the coil, the magnetic field at the centre of the coil is

1

2 π \ln (2) mT

2

\frac{π}{2} \ln (2) mT

3

π \ln (2) mT

4

\sqrt{2} π \ln (2) mT

Explanation:

A Given,
$N = 100$
$r_{1} = 1 cm = 1 \times 10^{- 2} m$
$r_{2} = 2 cm = 2 \times 10^{- 2} m$
$I = 1 A$
$B = \frac{μ_{0} NI}{2 (r_{2} - r_{1})} \ln (\frac{r_{2}}{r_{1}})$
$= \frac{4 π \times 10^{- 7} \times 100 \times 1}{2 \times (2 - 1) \times 10^{- 2}} \ln (\frac{2 \times 10^{- 2}}{1 \times 10^{- 2}})$
$B = 2 π \ln (2) \times 10^{- 3} T$
$B = 2 π \ln (2) mT$

TS- EAMCET-11.09.2020

Moving Charges & Magnetism

153180 The magnetic induction at point $O$ of the given infinitely long current carrying wire shown in the figure below is

1

\frac{μ_{0} I}{4 π R} (1 - \frac{3 π}{2})

2

\frac{μ_{0} I}{2 R (1 + π)}

3

\frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})

4

\frac{μ_{0} I}{4 π R}

Explanation:

C Magnetic field at point $O =$ Magnetic field at $O$ due to straight wire + Magnetic field of wire of loop
$B = \frac{μ_{0} I}{4 π R} + \frac{μ_{0} I}{4 π R} (\frac{3 π}{2})$
$B = \frac{μ_{0} I}{4 π R} + \frac{3}{8} \frac{μ_{0} I}{R}$
$B = \frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})$

Karnataka CET-2020

Moving Charges & Magnetism

153176 The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$ , is

1

\frac{μ_{0} I}{4 π r} T

2

\frac{μ_{o} I}{2 π r} T

3

\frac{μ_{o} I}{2 π R} T

4

\frac{μ_{o} I}{4 π R} T

Explanation:

B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $I$ by a conductor of Radius $R$ is given by
$B = \frac{μ_{0} I}{2 π r} T$

TS- EAMCET-09.09.2020

Moving Charges & Magnetism

153181 A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is

1 Zero

2

3 μ_{0} i / 32 R

, outward

3

3 μ_{0} i / 32 R

, inward

4

\frac{μ_{0} i}{2 R}

, inward

Explanation:

A Magnetic field due to $i_{1} = \frac{μ_{0} i_{1} θ_{1}}{2 R 2 π}$
$Magnetic field due to i_{2} = \frac{μ_{0} i_{2} θ_{2}}{2 R 2 π}$
For parallel combination-
$\frac{i_{1}}{i_{2}} = \frac{ρ l_{2}}{A} \times \frac{A}{ρ l_{1}} = \frac{l_{1}}{l_{2}}$
$\frac{i_{1}}{i_{2}} = \frac{\frac{1}{4} (2 π R)}{\frac{3}{4} (2 π R)}$
$i_{1} = \frac{i_{2}}{3}, i_{2} = 3 i_{1}$
Net magnetic field-
$= \frac{μ_{0} i_{1}}{2 R} (\frac{θ_{1}}{2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{θ_{2}}{2 π})$
$= \frac{μ_{0}}{2 R} (\frac{3 π}{2 \times 2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{π}{2 π}) \times \frac{1}{2}$
$= \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{i_{2}}{4}) = \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{3 i_{1}}{4}) = 0$

NEET Odisha-2019

Moving Charges & Magnetism

153179 A coil having 100 turns is wound tightly in the form of a spiral with inner and outer radii $1 cm$ and $2 cm$ , respectively. When a current $1 A$ passes through the coil, the magnetic field at the centre of the coil is

1

2 π \ln (2) mT

2

\frac{π}{2} \ln (2) mT

3

π \ln (2) mT

4

\sqrt{2} π \ln (2) mT

Explanation:

A Given,
$N = 100$
$r_{1} = 1 cm = 1 \times 10^{- 2} m$
$r_{2} = 2 cm = 2 \times 10^{- 2} m$
$I = 1 A$
$B = \frac{μ_{0} NI}{2 (r_{2} - r_{1})} \ln (\frac{r_{2}}{r_{1}})$
$= \frac{4 π \times 10^{- 7} \times 100 \times 1}{2 \times (2 - 1) \times 10^{- 2}} \ln (\frac{2 \times 10^{- 2}}{1 \times 10^{- 2}})$
$B = 2 π \ln (2) \times 10^{- 3} T$
$B = 2 π \ln (2) mT$

TS- EAMCET-11.09.2020

Moving Charges & Magnetism

153180 The magnetic induction at point $O$ of the given infinitely long current carrying wire shown in the figure below is

1

\frac{μ_{0} I}{4 π R} (1 - \frac{3 π}{2})

2

\frac{μ_{0} I}{2 R (1 + π)}

3

\frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})

4

\frac{μ_{0} I}{4 π R}

Explanation:

C Magnetic field at point $O =$ Magnetic field at $O$ due to straight wire + Magnetic field of wire of loop
$B = \frac{μ_{0} I}{4 π R} + \frac{μ_{0} I}{4 π R} (\frac{3 π}{2})$
$B = \frac{μ_{0} I}{4 π R} + \frac{3}{8} \frac{μ_{0} I}{R}$
$B = \frac{μ_{0} I}{4 π R} (1 + \frac{3 π}{2})$

Karnataka CET-2020

Moving Charges & Magnetism

153176 The magnetic field at point $P$ of given figure due to carrying of current $I$ by a conductor of radius $R$ , is

1

\frac{μ_{0} I}{4 π r} T

2

\frac{μ_{o} I}{2 π r} T

3

\frac{μ_{o} I}{2 π R} T

4

\frac{μ_{o} I}{4 π R} T

Explanation:

B The magnetic field at point $P$ distance $r$ of given figure due to carrying current $I$ by a conductor of Radius $R$ is given by
$B = \frac{μ_{0} I}{2 π r} T$

TS- EAMCET-09.09.2020

Moving Charges & Magnetism

153181 A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is

1 Zero

2

3 μ_{0} i / 32 R

, outward

3

3 μ_{0} i / 32 R

, inward

4

\frac{μ_{0} i}{2 R}

, inward

Explanation:

A Magnetic field due to $i_{1} = \frac{μ_{0} i_{1} θ_{1}}{2 R 2 π}$
$Magnetic field due to i_{2} = \frac{μ_{0} i_{2} θ_{2}}{2 R 2 π}$
For parallel combination-
$\frac{i_{1}}{i_{2}} = \frac{ρ l_{2}}{A} \times \frac{A}{ρ l_{1}} = \frac{l_{1}}{l_{2}}$
$\frac{i_{1}}{i_{2}} = \frac{\frac{1}{4} (2 π R)}{\frac{3}{4} (2 π R)}$
$i_{1} = \frac{i_{2}}{3}, i_{2} = 3 i_{1}$
Net magnetic field-
$= \frac{μ_{0} i_{1}}{2 R} (\frac{θ_{1}}{2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{θ_{2}}{2 π})$
$= \frac{μ_{0}}{2 R} (\frac{3 π}{2 \times 2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{π}{2 π}) \times \frac{1}{2}$
$= \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{i_{2}}{4}) = \frac{μ_{0}}{2 R} (\frac{3 i_{1}}{4} - \frac{3 i_{1}}{4}) = 0$

NEET Odisha-2019

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: