QBManager

PHXII04:MOVING CHARGES AND MAGNETISM

363024 The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_{A} = 10 c m$ and $r_{B} = 20 c m$ respectively are equal if (Where $N_{A}, I_{A}$ and $N_{B}, I_{B}$ are number of turn and current of $A$ and $B$ respectively)

1

N_{A} = 2 N_{B}

2

2 N_{A} I_{A} = N_{B} I_{B}

3

N_{A} I_{A} = 4 N_{B} I_{B}

4

4 N_{A} I_{A} = N_{B} I_{B}

Explanation:

Magnetic moment of the current carrying loop is
$M = N I A$
$M \propto R^{2}$
$(∵ A = π R^{2})$
Here, $\frac{R_{A}}{R_{B}} = \frac{1}{2}$ .
Therefore, $\frac{M_{A}}{M_{B}} = \frac{R_{A}^{2}}{R_{B}^{2}} = \frac{1}{4} = 1$

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

363025 Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the effective magnetic moment will be

1

π : 2

2

2 : π

3

π : 4

4

4 : π

Explanation:

Suppose length of each wire is $l$ .
$A_{square} = {(\frac{l}{4})}^{2} = \frac{l^{2}}{16}$
supporting img

$A_{cirde} = π r^{2} = π {(\frac{l}{2 π})}^{2} = \frac{l^{2}}{4 π}$
$∵$ Magnetic moment $M = i A$
$\Rightarrow \frac{M_{square}}{M_{circle}} = \frac{A_{square}}{A_{circle}} = \frac{l^{2} / 16}{l^{2} / 4 π} = \frac{π}{4}$

PHXII04:MOVING CHARGES AND MAGNETISM

363026 The magnetic dipole moment of a current loop is independent of

1 Magnetic field in which it is lying

2 Number of turns

3 Area of the loop

4 Current in the loop

Explanation:

Current loop acts as a magnetic dipole.
Its magnetic moment is given by, $M = NIA$ where $N =$ number of turns, $I =$ current in a loop, $A =$ area of the loop.
From the relation, we can conclude that magnetic dipole moment of a current loop is independent of magnetic field in which it is lying.

KCET - 2009

PHXII04:MOVING CHARGES AND MAGNETISM

363027 A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in $M K S$ unit is $M$ , the length of wire will be

1

\frac{4 π}{M}

2

\sqrt{\frac{4 π M}{i}}

3

\sqrt{\frac{4 π i}{M}}

4

\frac{M π}{i}

Explanation:

Magnetic moment of a circular loop carrying current $i$ is given by
$M = i A = i π r^{2} or r = \sqrt{\frac{M}{π i}}$
Length of wire, $I = 2 π r = 2 π \sqrt{\frac{M}{π i}} = \sqrt{\frac{4 π M}{i}}$

PHXII04:MOVING CHARGES AND MAGNETISM

363024 The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_{A} = 10 c m$ and $r_{B} = 20 c m$ respectively are equal if (Where $N_{A}, I_{A}$ and $N_{B}, I_{B}$ are number of turn and current of $A$ and $B$ respectively)

1

N_{A} = 2 N_{B}

2

2 N_{A} I_{A} = N_{B} I_{B}

3

N_{A} I_{A} = 4 N_{B} I_{B}

4

4 N_{A} I_{A} = N_{B} I_{B}

Explanation:

Magnetic moment of the current carrying loop is
$M = N I A$
$M \propto R^{2}$
$(∵ A = π R^{2})$
Here, $\frac{R_{A}}{R_{B}} = \frac{1}{2}$ .
Therefore, $\frac{M_{A}}{M_{B}} = \frac{R_{A}^{2}}{R_{B}^{2}} = \frac{1}{4} = 1$

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

363025 Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the effective magnetic moment will be

1

π : 2

2

2 : π

3

π : 4

4

4 : π

Explanation:

Suppose length of each wire is $l$ .
$A_{square} = {(\frac{l}{4})}^{2} = \frac{l^{2}}{16}$
supporting img

$A_{cirde} = π r^{2} = π {(\frac{l}{2 π})}^{2} = \frac{l^{2}}{4 π}$
$∵$ Magnetic moment $M = i A$
$\Rightarrow \frac{M_{square}}{M_{circle}} = \frac{A_{square}}{A_{circle}} = \frac{l^{2} / 16}{l^{2} / 4 π} = \frac{π}{4}$

PHXII04:MOVING CHARGES AND MAGNETISM

363026 The magnetic dipole moment of a current loop is independent of

1 Magnetic field in which it is lying

2 Number of turns

3 Area of the loop

4 Current in the loop

Explanation:

Current loop acts as a magnetic dipole.
Its magnetic moment is given by, $M = NIA$ where $N =$ number of turns, $I =$ current in a loop, $A =$ area of the loop.
From the relation, we can conclude that magnetic dipole moment of a current loop is independent of magnetic field in which it is lying.

KCET - 2009

PHXII04:MOVING CHARGES AND MAGNETISM

363027 A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in $M K S$ unit is $M$ , the length of wire will be

1

\frac{4 π}{M}

2

\sqrt{\frac{4 π M}{i}}

3

\sqrt{\frac{4 π i}{M}}

4

\frac{M π}{i}

Explanation:

Magnetic moment of a circular loop carrying current $i$ is given by
$M = i A = i π r^{2} or r = \sqrt{\frac{M}{π i}}$
Length of wire, $I = 2 π r = 2 π \sqrt{\frac{M}{π i}} = \sqrt{\frac{4 π M}{i}}$

PHXII04:MOVING CHARGES AND MAGNETISM

363024 The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_{A} = 10 c m$ and $r_{B} = 20 c m$ respectively are equal if (Where $N_{A}, I_{A}$ and $N_{B}, I_{B}$ are number of turn and current of $A$ and $B$ respectively)

1

N_{A} = 2 N_{B}

2

2 N_{A} I_{A} = N_{B} I_{B}

3

N_{A} I_{A} = 4 N_{B} I_{B}

4

4 N_{A} I_{A} = N_{B} I_{B}

Explanation:

Magnetic moment of the current carrying loop is
$M = N I A$
$M \propto R^{2}$
$(∵ A = π R^{2})$
Here, $\frac{R_{A}}{R_{B}} = \frac{1}{2}$ .
Therefore, $\frac{M_{A}}{M_{B}} = \frac{R_{A}^{2}}{R_{B}^{2}} = \frac{1}{4} = 1$

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

363025 Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the effective magnetic moment will be

1

π : 2

2

2 : π

3

π : 4

4

4 : π

Explanation:

Suppose length of each wire is $l$ .
$A_{square} = {(\frac{l}{4})}^{2} = \frac{l^{2}}{16}$
supporting img

$A_{cirde} = π r^{2} = π {(\frac{l}{2 π})}^{2} = \frac{l^{2}}{4 π}$
$∵$ Magnetic moment $M = i A$
$\Rightarrow \frac{M_{square}}{M_{circle}} = \frac{A_{square}}{A_{circle}} = \frac{l^{2} / 16}{l^{2} / 4 π} = \frac{π}{4}$

PHXII04:MOVING CHARGES AND MAGNETISM

363026 The magnetic dipole moment of a current loop is independent of

1 Magnetic field in which it is lying

2 Number of turns

3 Area of the loop

4 Current in the loop

Explanation:

Current loop acts as a magnetic dipole.
Its magnetic moment is given by, $M = NIA$ where $N =$ number of turns, $I =$ current in a loop, $A =$ area of the loop.
From the relation, we can conclude that magnetic dipole moment of a current loop is independent of magnetic field in which it is lying.

KCET - 2009

PHXII04:MOVING CHARGES AND MAGNETISM

363027 A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in $M K S$ unit is $M$ , the length of wire will be

1

\frac{4 π}{M}

2

\sqrt{\frac{4 π M}{i}}

3

\sqrt{\frac{4 π i}{M}}

4

\frac{M π}{i}

Explanation:

Magnetic moment of a circular loop carrying current $i$ is given by
$M = i A = i π r^{2} or r = \sqrt{\frac{M}{π i}}$
Length of wire, $I = 2 π r = 2 π \sqrt{\frac{M}{π i}} = \sqrt{\frac{4 π M}{i}}$

PHXII04:MOVING CHARGES AND MAGNETISM

363024 The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_{A} = 10 c m$ and $r_{B} = 20 c m$ respectively are equal if (Where $N_{A}, I_{A}$ and $N_{B}, I_{B}$ are number of turn and current of $A$ and $B$ respectively)

1

N_{A} = 2 N_{B}

2

2 N_{A} I_{A} = N_{B} I_{B}

3

N_{A} I_{A} = 4 N_{B} I_{B}

4

4 N_{A} I_{A} = N_{B} I_{B}

Explanation:

Magnetic moment of the current carrying loop is
$M = N I A$
$M \propto R^{2}$
$(∵ A = π R^{2})$
Here, $\frac{R_{A}}{R_{B}} = \frac{1}{2}$ .
Therefore, $\frac{M_{A}}{M_{B}} = \frac{R_{A}^{2}}{R_{B}^{2}} = \frac{1}{4} = 1$

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

363025 Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the effective magnetic moment will be

1

π : 2

2

2 : π

3

π : 4

4

4 : π

Explanation:

Suppose length of each wire is $l$ .
$A_{square} = {(\frac{l}{4})}^{2} = \frac{l^{2}}{16}$
supporting img

$A_{cirde} = π r^{2} = π {(\frac{l}{2 π})}^{2} = \frac{l^{2}}{4 π}$
$∵$ Magnetic moment $M = i A$
$\Rightarrow \frac{M_{square}}{M_{circle}} = \frac{A_{square}}{A_{circle}} = \frac{l^{2} / 16}{l^{2} / 4 π} = \frac{π}{4}$

PHXII04:MOVING CHARGES AND MAGNETISM

363026 The magnetic dipole moment of a current loop is independent of

1 Magnetic field in which it is lying

2 Number of turns

3 Area of the loop

4 Current in the loop

Explanation:

Current loop acts as a magnetic dipole.
Its magnetic moment is given by, $M = NIA$ where $N =$ number of turns, $I =$ current in a loop, $A =$ area of the loop.
From the relation, we can conclude that magnetic dipole moment of a current loop is independent of magnetic field in which it is lying.

KCET - 2009

PHXII04:MOVING CHARGES AND MAGNETISM

363027 A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in $M K S$ unit is $M$ , the length of wire will be

1

\frac{4 π}{M}

2

\sqrt{\frac{4 π M}{i}}

3

\sqrt{\frac{4 π i}{M}}

4

\frac{M π}{i}

Explanation:

Magnetic moment of a circular loop carrying current $i$ is given by
$M = i A = i π r^{2} or r = \sqrt{\frac{M}{π i}}$
Length of wire, $I = 2 π r = 2 π \sqrt{\frac{M}{π i}} = \sqrt{\frac{4 π M}{i}}$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: