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PHXII04:MOVING CHARGES AND MAGNETISM

362570 An arc of a circle of radius $R$ subtends an angle $π / 2$ at the centre. It carries a current $i$ . The magnetic field at the centre will be

1

\frac{μ_{0} i}{4 R}

2

\frac{μ_{0} i}{8 R}

3

\frac{μ_{0} i}{2 R}

4

\frac{2 μ_{0} i}{5 R}

Explanation:

$B = \frac{μ_{0}}{4 π} \frac{θ i}{r} = \frac{μ_{0}}{4 π} \times \frac{π}{2} \times \frac{1}{R} = \frac{μ_{0} i}{8 R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362571 Magnetic field at the centre of a circular loop of area $A$ is $B$ . The magnetic moment of the loop will be

1

\frac{B A^{2}}{μ_{0} π}

2

\frac{B A^{3 / 2}}{μ_{0} π}

3

\frac{B A^{3 / 2}}{μ_{0} π^{1 / 2}}

4

\frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}

Explanation:

Magnetic field at the centre of the circular loop is given by
$B = \frac{μ_{0} I}{2 r}; I = \frac{2 B r}{μ_{0}}$
Also, $A = π r^{2}; r = \sqrt{\frac{A}{π}}$
Magnetic moment of the loop, $M = I A$
or, $M = \frac{2 B r}{μ_{0}} A = \frac{2 B}{μ_{0}} \sqrt{\frac{A}{π}} A$
$M = \frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}$

PHXII04:MOVING CHARGES AND MAGNETISM

362572 $A$ and $B$ are the two concentric circular conductors of center $O$ and carrying currents $I_{1}$ and $I_{2}$ as shown in the adjacent figure. If ratio of their radii is $1 : 2$ and ratio of the flux densities at $O$ due to $A$ and $B$ is $1 : 3$ then the value of $I_{1} /$ $I_{2}$ is
supporting img

1

1 / 6

2

1 / 4

3

1 / 3

4

1 / 2

Explanation:

$r_{1} : r_{2} = 1 : 2$ and $B_{1} : B_{2} = 1 : 3$ (flux density)
$B = \frac{μ_{0}}{4 π} \frac{2 π n I}{R}$
$\frac{I_{1}}{I_{2}} = \frac{B_{1}}{B_{2}} \frac{R_{1}}{R_{2}} = \frac{1 \times 1}{3 \times 2} = \frac{1}{6} \frac{I_{1}}{I_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362573 If a current $I$ is flowing in a loop of radius $r$ as shown in figure, then the magnetic field induction at the centre $O$ will be
supporting img

1 zero

2

\frac{μ_{0} I θ}{4 π r}

3

\frac{μ_{0} I \sin θ}{4 π r}

4

\frac{2 μ_{0} I \sin θ}{4 π r^{2}}

Explanation:

supporting img
The magnetic field induction at the centre $O$ is
$B = \frac{μ_{0} I θ}{4 π r}$

PHXII04:MOVING CHARGES AND MAGNETISM

362570 An arc of a circle of radius $R$ subtends an angle $π / 2$ at the centre. It carries a current $i$ . The magnetic field at the centre will be

1

\frac{μ_{0} i}{4 R}

2

\frac{μ_{0} i}{8 R}

3

\frac{μ_{0} i}{2 R}

4

\frac{2 μ_{0} i}{5 R}

Explanation:

$B = \frac{μ_{0}}{4 π} \frac{θ i}{r} = \frac{μ_{0}}{4 π} \times \frac{π}{2} \times \frac{1}{R} = \frac{μ_{0} i}{8 R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362571 Magnetic field at the centre of a circular loop of area $A$ is $B$ . The magnetic moment of the loop will be

1

\frac{B A^{2}}{μ_{0} π}

2

\frac{B A^{3 / 2}}{μ_{0} π}

3

\frac{B A^{3 / 2}}{μ_{0} π^{1 / 2}}

4

\frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}

Explanation:

Magnetic field at the centre of the circular loop is given by
$B = \frac{μ_{0} I}{2 r}; I = \frac{2 B r}{μ_{0}}$
Also, $A = π r^{2}; r = \sqrt{\frac{A}{π}}$
Magnetic moment of the loop, $M = I A$
or, $M = \frac{2 B r}{μ_{0}} A = \frac{2 B}{μ_{0}} \sqrt{\frac{A}{π}} A$
$M = \frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}$

PHXII04:MOVING CHARGES AND MAGNETISM

362572 $A$ and $B$ are the two concentric circular conductors of center $O$ and carrying currents $I_{1}$ and $I_{2}$ as shown in the adjacent figure. If ratio of their radii is $1 : 2$ and ratio of the flux densities at $O$ due to $A$ and $B$ is $1 : 3$ then the value of $I_{1} /$ $I_{2}$ is
supporting img

1

1 / 6

2

1 / 4

3

1 / 3

4

1 / 2

Explanation:

$r_{1} : r_{2} = 1 : 2$ and $B_{1} : B_{2} = 1 : 3$ (flux density)
$B = \frac{μ_{0}}{4 π} \frac{2 π n I}{R}$
$\frac{I_{1}}{I_{2}} = \frac{B_{1}}{B_{2}} \frac{R_{1}}{R_{2}} = \frac{1 \times 1}{3 \times 2} = \frac{1}{6} \frac{I_{1}}{I_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362573 If a current $I$ is flowing in a loop of radius $r$ as shown in figure, then the magnetic field induction at the centre $O$ will be
supporting img

1 zero

2

\frac{μ_{0} I θ}{4 π r}

3

\frac{μ_{0} I \sin θ}{4 π r}

4

\frac{2 μ_{0} I \sin θ}{4 π r^{2}}

Explanation:

supporting img
The magnetic field induction at the centre $O$ is
$B = \frac{μ_{0} I θ}{4 π r}$

PHXII04:MOVING CHARGES AND MAGNETISM

362570 An arc of a circle of radius $R$ subtends an angle $π / 2$ at the centre. It carries a current $i$ . The magnetic field at the centre will be

1

\frac{μ_{0} i}{4 R}

2

\frac{μ_{0} i}{8 R}

3

\frac{μ_{0} i}{2 R}

4

\frac{2 μ_{0} i}{5 R}

Explanation:

$B = \frac{μ_{0}}{4 π} \frac{θ i}{r} = \frac{μ_{0}}{4 π} \times \frac{π}{2} \times \frac{1}{R} = \frac{μ_{0} i}{8 R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362571 Magnetic field at the centre of a circular loop of area $A$ is $B$ . The magnetic moment of the loop will be

1

\frac{B A^{2}}{μ_{0} π}

2

\frac{B A^{3 / 2}}{μ_{0} π}

3

\frac{B A^{3 / 2}}{μ_{0} π^{1 / 2}}

4

\frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}

Explanation:

Magnetic field at the centre of the circular loop is given by
$B = \frac{μ_{0} I}{2 r}; I = \frac{2 B r}{μ_{0}}$
Also, $A = π r^{2}; r = \sqrt{\frac{A}{π}}$
Magnetic moment of the loop, $M = I A$
or, $M = \frac{2 B r}{μ_{0}} A = \frac{2 B}{μ_{0}} \sqrt{\frac{A}{π}} A$
$M = \frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}$

PHXII04:MOVING CHARGES AND MAGNETISM

362572 $A$ and $B$ are the two concentric circular conductors of center $O$ and carrying currents $I_{1}$ and $I_{2}$ as shown in the adjacent figure. If ratio of their radii is $1 : 2$ and ratio of the flux densities at $O$ due to $A$ and $B$ is $1 : 3$ then the value of $I_{1} /$ $I_{2}$ is
supporting img

1

1 / 6

2

1 / 4

3

1 / 3

4

1 / 2

Explanation:

$r_{1} : r_{2} = 1 : 2$ and $B_{1} : B_{2} = 1 : 3$ (flux density)
$B = \frac{μ_{0}}{4 π} \frac{2 π n I}{R}$
$\frac{I_{1}}{I_{2}} = \frac{B_{1}}{B_{2}} \frac{R_{1}}{R_{2}} = \frac{1 \times 1}{3 \times 2} = \frac{1}{6} \frac{I_{1}}{I_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362573 If a current $I$ is flowing in a loop of radius $r$ as shown in figure, then the magnetic field induction at the centre $O$ will be
supporting img

1 zero

2

\frac{μ_{0} I θ}{4 π r}

3

\frac{μ_{0} I \sin θ}{4 π r}

4

\frac{2 μ_{0} I \sin θ}{4 π r^{2}}

Explanation:

supporting img
The magnetic field induction at the centre $O$ is
$B = \frac{μ_{0} I θ}{4 π r}$

PHXII04:MOVING CHARGES AND MAGNETISM

362570 An arc of a circle of radius $R$ subtends an angle $π / 2$ at the centre. It carries a current $i$ . The magnetic field at the centre will be

1

\frac{μ_{0} i}{4 R}

2

\frac{μ_{0} i}{8 R}

3

\frac{μ_{0} i}{2 R}

4

\frac{2 μ_{0} i}{5 R}

Explanation:

$B = \frac{μ_{0}}{4 π} \frac{θ i}{r} = \frac{μ_{0}}{4 π} \times \frac{π}{2} \times \frac{1}{R} = \frac{μ_{0} i}{8 R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362571 Magnetic field at the centre of a circular loop of area $A$ is $B$ . The magnetic moment of the loop will be

1

\frac{B A^{2}}{μ_{0} π}

2

\frac{B A^{3 / 2}}{μ_{0} π}

3

\frac{B A^{3 / 2}}{μ_{0} π^{1 / 2}}

4

\frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}

Explanation:

Magnetic field at the centre of the circular loop is given by
$B = \frac{μ_{0} I}{2 r}; I = \frac{2 B r}{μ_{0}}$
Also, $A = π r^{2}; r = \sqrt{\frac{A}{π}}$
Magnetic moment of the loop, $M = I A$
or, $M = \frac{2 B r}{μ_{0}} A = \frac{2 B}{μ_{0}} \sqrt{\frac{A}{π}} A$
$M = \frac{2 B A^{3 / 2}}{μ_{0} π^{1 / 2}}$

PHXII04:MOVING CHARGES AND MAGNETISM

362572 $A$ and $B$ are the two concentric circular conductors of center $O$ and carrying currents $I_{1}$ and $I_{2}$ as shown in the adjacent figure. If ratio of their radii is $1 : 2$ and ratio of the flux densities at $O$ due to $A$ and $B$ is $1 : 3$ then the value of $I_{1} /$ $I_{2}$ is
supporting img

1

1 / 6

2

1 / 4

3

1 / 3

4

1 / 2

Explanation:

$r_{1} : r_{2} = 1 : 2$ and $B_{1} : B_{2} = 1 : 3$ (flux density)
$B = \frac{μ_{0}}{4 π} \frac{2 π n I}{R}$
$\frac{I_{1}}{I_{2}} = \frac{B_{1}}{B_{2}} \frac{R_{1}}{R_{2}} = \frac{1 \times 1}{3 \times 2} = \frac{1}{6} \frac{I_{1}}{I_{2}} = \frac{1}{6}$

PHXII04:MOVING CHARGES AND MAGNETISM

362573 If a current $I$ is flowing in a loop of radius $r$ as shown in figure, then the magnetic field induction at the centre $O$ will be
supporting img

1 zero

2

\frac{μ_{0} I θ}{4 π r}

3

\frac{μ_{0} I \sin θ}{4 π r}

4

\frac{2 μ_{0} I \sin θ}{4 π r^{2}}

Explanation:

supporting img
The magnetic field induction at the centre $O$ is
$B = \frac{μ_{0} I θ}{4 π r}$

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