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PHXII04:MOVING CHARGES AND MAGNETISM

362609 A square conducting loop of side length $L$ carries a current $I$ . The magnetic field at the centre of the loop is

1 Linearly proportional to

L

2 Independent to

L

3 Inversely proportional to

L

4 Proportional to

L^{2}

Explanation:

Magnetic field at the centre of due to either arms
$\begin{aligned} B_{1} = \frac{μ_{0}}{4 π} \times \frac{1}{(L / 2)} [\sin 45^{\circ} + \sin 45^{\circ}] \\ = \frac{μ_{0}}{4 π} \times \frac{2 \sqrt{2} I}{L} \end{aligned}$
Field at centre due to the four arms of the square
$B = 4 B_{1} = \frac{μ_{0}}{π} \times \frac{2 \sqrt{2} I}{L}$
i.e., $B \propto \frac{1}{L}$

PHXII04:MOVING CHARGES AND MAGNETISM

362610 The magnetic field at the centre of square of side $a$ is
supporting img

1

\frac{\sqrt{2} μ_{0}}{π a}

2

\frac{\sqrt{2} μ_{0} I}{3 π a}

3

\frac{2}{3} \frac{μ_{0} I}{a}

4 zero

Explanation:

Magnetic field,
$B = 2 [\begin{array}{l} \frac{μ_{0}}{4 π} \frac{2 I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) - \\ \frac{μ_{0}}{4 π} \cdot \frac{I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) \end{array}]$
$= \frac{\sqrt{2} μ_{0} I}{3 π a}$
supporting img

PHXII04:MOVING CHARGES AND MAGNETISM

362611 Calculate the magnetic field at the centre of a coil in the form of a square of side 2 a carrying a current $I$ .

1

\frac{2 μ_{0} I}{π a}

2

\frac{\sqrt{2} μ_{0} I}{π a}

3

\frac{5 μ_{0} I}{π a}

4

\frac{\sqrt{3} μ_{0} I}{π a}

Explanation:

The current carrying coil $A B C D$ may be assumed to be made of four current-carrying conductors $A B, B C, C D$ and $D A$ .Magnetic field at $O$ due to current-carrying conductor $A B$ is :
$B = \frac{μ_{0} I}{4 π a} [\sin 45^{\circ} + \sin 45^{\circ}]$
$= \frac{μ_{0} I}{4 π a} 2 \sin 45^{\circ}$
or $B = \frac{μ_{0} I}{4 π a} \times 2 \times \frac{1}{\sqrt{2}}$
or $B = \frac{\sqrt{2} μ_{0} I}{4 π a}$
supporting img
Total magnetic field at $O$
$B^{'} = 4 B = 4 \times \frac{\sqrt{2} μ_{0} I}{4 π a} = \frac{\sqrt{2} μ_{0} I}{π a}$

PHXII04:MOVING CHARGES AND MAGNETISM

362612 An equilateral triangle is made by uniform wires $A B$ , $B C$ , $C A$ . A current $I$ enters at $A$ and leaves from the mid point of $B C$ . If the lengths of each side of the triangle is $L$ . The magnetic field $B$ at the centroid $O$ of the triangle is
supporting img

1

\frac{μ_{0}}{4 π} (\frac{4 I}{L})

2

\frac{μ_{0}}{4 π} (\frac{2 I}{L})

3

\frac{μ_{0}}{2 π} (\frac{4 I}{L})

4 Zero

Explanation:

supporting img
$\begin{aligned} {\vec{B}}_{1} + {\vec{B}}_{2} = 0 \\ B_{1} \to due to left part \\ B_{2} \to due right part \end{aligned}$
$B_{1}, B_{2}$ at centroid are numerically equal & opposite to each other. (i,e. one is into the paper and the other is towards the reader)

PHXII04:MOVING CHARGES AND MAGNETISM

362609 A square conducting loop of side length $L$ carries a current $I$ . The magnetic field at the centre of the loop is

1 Linearly proportional to

L

2 Independent to

L

3 Inversely proportional to

L

4 Proportional to

L^{2}

Explanation:

Magnetic field at the centre of due to either arms
$\begin{aligned} B_{1} = \frac{μ_{0}}{4 π} \times \frac{1}{(L / 2)} [\sin 45^{\circ} + \sin 45^{\circ}] \\ = \frac{μ_{0}}{4 π} \times \frac{2 \sqrt{2} I}{L} \end{aligned}$
Field at centre due to the four arms of the square
$B = 4 B_{1} = \frac{μ_{0}}{π} \times \frac{2 \sqrt{2} I}{L}$
i.e., $B \propto \frac{1}{L}$

PHXII04:MOVING CHARGES AND MAGNETISM

362610 The magnetic field at the centre of square of side $a$ is
supporting img

1

\frac{\sqrt{2} μ_{0}}{π a}

2

\frac{\sqrt{2} μ_{0} I}{3 π a}

3

\frac{2}{3} \frac{μ_{0} I}{a}

4 zero

Explanation:

Magnetic field,
$B = 2 [\begin{array}{l} \frac{μ_{0}}{4 π} \frac{2 I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) - \\ \frac{μ_{0}}{4 π} \cdot \frac{I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) \end{array}]$
$= \frac{\sqrt{2} μ_{0} I}{3 π a}$
supporting img

PHXII04:MOVING CHARGES AND MAGNETISM

362611 Calculate the magnetic field at the centre of a coil in the form of a square of side 2 a carrying a current $I$ .

1

\frac{2 μ_{0} I}{π a}

2

\frac{\sqrt{2} μ_{0} I}{π a}

3

\frac{5 μ_{0} I}{π a}

4

\frac{\sqrt{3} μ_{0} I}{π a}

Explanation:

The current carrying coil $A B C D$ may be assumed to be made of four current-carrying conductors $A B, B C, C D$ and $D A$ .Magnetic field at $O$ due to current-carrying conductor $A B$ is :
$B = \frac{μ_{0} I}{4 π a} [\sin 45^{\circ} + \sin 45^{\circ}]$
$= \frac{μ_{0} I}{4 π a} 2 \sin 45^{\circ}$
or $B = \frac{μ_{0} I}{4 π a} \times 2 \times \frac{1}{\sqrt{2}}$
or $B = \frac{\sqrt{2} μ_{0} I}{4 π a}$
supporting img
Total magnetic field at $O$
$B^{'} = 4 B = 4 \times \frac{\sqrt{2} μ_{0} I}{4 π a} = \frac{\sqrt{2} μ_{0} I}{π a}$

PHXII04:MOVING CHARGES AND MAGNETISM

362612 An equilateral triangle is made by uniform wires $A B$ , $B C$ , $C A$ . A current $I$ enters at $A$ and leaves from the mid point of $B C$ . If the lengths of each side of the triangle is $L$ . The magnetic field $B$ at the centroid $O$ of the triangle is
supporting img

1

\frac{μ_{0}}{4 π} (\frac{4 I}{L})

2

\frac{μ_{0}}{4 π} (\frac{2 I}{L})

3

\frac{μ_{0}}{2 π} (\frac{4 I}{L})

4 Zero

Explanation:

supporting img
$\begin{aligned} {\vec{B}}_{1} + {\vec{B}}_{2} = 0 \\ B_{1} \to due to left part \\ B_{2} \to due right part \end{aligned}$
$B_{1}, B_{2}$ at centroid are numerically equal & opposite to each other. (i,e. one is into the paper and the other is towards the reader)

PHXII04:MOVING CHARGES AND MAGNETISM

362609 A square conducting loop of side length $L$ carries a current $I$ . The magnetic field at the centre of the loop is

1 Linearly proportional to

L

2 Independent to

L

3 Inversely proportional to

L

4 Proportional to

L^{2}

Explanation:

Magnetic field at the centre of due to either arms
$\begin{aligned} B_{1} = \frac{μ_{0}}{4 π} \times \frac{1}{(L / 2)} [\sin 45^{\circ} + \sin 45^{\circ}] \\ = \frac{μ_{0}}{4 π} \times \frac{2 \sqrt{2} I}{L} \end{aligned}$
Field at centre due to the four arms of the square
$B = 4 B_{1} = \frac{μ_{0}}{π} \times \frac{2 \sqrt{2} I}{L}$
i.e., $B \propto \frac{1}{L}$

PHXII04:MOVING CHARGES AND MAGNETISM

362610 The magnetic field at the centre of square of side $a$ is
supporting img

1

\frac{\sqrt{2} μ_{0}}{π a}

2

\frac{\sqrt{2} μ_{0} I}{3 π a}

3

\frac{2}{3} \frac{μ_{0} I}{a}

4 zero

Explanation:

Magnetic field,
$B = 2 [\begin{array}{l} \frac{μ_{0}}{4 π} \frac{2 I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) - \\ \frac{μ_{0}}{4 π} \cdot \frac{I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) \end{array}]$
$= \frac{\sqrt{2} μ_{0} I}{3 π a}$
supporting img

PHXII04:MOVING CHARGES AND MAGNETISM

362611 Calculate the magnetic field at the centre of a coil in the form of a square of side 2 a carrying a current $I$ .

1

\frac{2 μ_{0} I}{π a}

2

\frac{\sqrt{2} μ_{0} I}{π a}

3

\frac{5 μ_{0} I}{π a}

4

\frac{\sqrt{3} μ_{0} I}{π a}

Explanation:

The current carrying coil $A B C D$ may be assumed to be made of four current-carrying conductors $A B, B C, C D$ and $D A$ .Magnetic field at $O$ due to current-carrying conductor $A B$ is :
$B = \frac{μ_{0} I}{4 π a} [\sin 45^{\circ} + \sin 45^{\circ}]$
$= \frac{μ_{0} I}{4 π a} 2 \sin 45^{\circ}$
or $B = \frac{μ_{0} I}{4 π a} \times 2 \times \frac{1}{\sqrt{2}}$
or $B = \frac{\sqrt{2} μ_{0} I}{4 π a}$
supporting img
Total magnetic field at $O$
$B^{'} = 4 B = 4 \times \frac{\sqrt{2} μ_{0} I}{4 π a} = \frac{\sqrt{2} μ_{0} I}{π a}$

PHXII04:MOVING CHARGES AND MAGNETISM

362612 An equilateral triangle is made by uniform wires $A B$ , $B C$ , $C A$ . A current $I$ enters at $A$ and leaves from the mid point of $B C$ . If the lengths of each side of the triangle is $L$ . The magnetic field $B$ at the centroid $O$ of the triangle is
supporting img

1

\frac{μ_{0}}{4 π} (\frac{4 I}{L})

2

\frac{μ_{0}}{4 π} (\frac{2 I}{L})

3

\frac{μ_{0}}{2 π} (\frac{4 I}{L})

4 Zero

Explanation:

supporting img
$\begin{aligned} {\vec{B}}_{1} + {\vec{B}}_{2} = 0 \\ B_{1} \to due to left part \\ B_{2} \to due right part \end{aligned}$
$B_{1}, B_{2}$ at centroid are numerically equal & opposite to each other. (i,e. one is into the paper and the other is towards the reader)

PHXII04:MOVING CHARGES AND MAGNETISM

362609 A square conducting loop of side length $L$ carries a current $I$ . The magnetic field at the centre of the loop is

1 Linearly proportional to

L

2 Independent to

L

3 Inversely proportional to

L

4 Proportional to

L^{2}

Explanation:

Magnetic field at the centre of due to either arms
$\begin{aligned} B_{1} = \frac{μ_{0}}{4 π} \times \frac{1}{(L / 2)} [\sin 45^{\circ} + \sin 45^{\circ}] \\ = \frac{μ_{0}}{4 π} \times \frac{2 \sqrt{2} I}{L} \end{aligned}$
Field at centre due to the four arms of the square
$B = 4 B_{1} = \frac{μ_{0}}{π} \times \frac{2 \sqrt{2} I}{L}$
i.e., $B \propto \frac{1}{L}$

PHXII04:MOVING CHARGES AND MAGNETISM

362610 The magnetic field at the centre of square of side $a$ is
supporting img

1

\frac{\sqrt{2} μ_{0}}{π a}

2

\frac{\sqrt{2} μ_{0} I}{3 π a}

3

\frac{2}{3} \frac{μ_{0} I}{a}

4 zero

Explanation:

Magnetic field,
$B = 2 [\begin{array}{l} \frac{μ_{0}}{4 π} \frac{2 I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) - \\ \frac{μ_{0}}{4 π} \cdot \frac{I / 3}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ}) \end{array}]$
$= \frac{\sqrt{2} μ_{0} I}{3 π a}$
supporting img

PHXII04:MOVING CHARGES AND MAGNETISM

362611 Calculate the magnetic field at the centre of a coil in the form of a square of side 2 a carrying a current $I$ .

1

\frac{2 μ_{0} I}{π a}

2

\frac{\sqrt{2} μ_{0} I}{π a}

3

\frac{5 μ_{0} I}{π a}

4

\frac{\sqrt{3} μ_{0} I}{π a}

Explanation:

The current carrying coil $A B C D$ may be assumed to be made of four current-carrying conductors $A B, B C, C D$ and $D A$ .Magnetic field at $O$ due to current-carrying conductor $A B$ is :
$B = \frac{μ_{0} I}{4 π a} [\sin 45^{\circ} + \sin 45^{\circ}]$
$= \frac{μ_{0} I}{4 π a} 2 \sin 45^{\circ}$
or $B = \frac{μ_{0} I}{4 π a} \times 2 \times \frac{1}{\sqrt{2}}$
or $B = \frac{\sqrt{2} μ_{0} I}{4 π a}$
supporting img
Total magnetic field at $O$
$B^{'} = 4 B = 4 \times \frac{\sqrt{2} μ_{0} I}{4 π a} = \frac{\sqrt{2} μ_{0} I}{π a}$

PHXII04:MOVING CHARGES AND MAGNETISM

362612 An equilateral triangle is made by uniform wires $A B$ , $B C$ , $C A$ . A current $I$ enters at $A$ and leaves from the mid point of $B C$ . If the lengths of each side of the triangle is $L$ . The magnetic field $B$ at the centroid $O$ of the triangle is
supporting img

1

\frac{μ_{0}}{4 π} (\frac{4 I}{L})

2

\frac{μ_{0}}{4 π} (\frac{2 I}{L})

3

\frac{μ_{0}}{2 π} (\frac{4 I}{L})

4 Zero

Explanation:

supporting img
$\begin{aligned} {\vec{B}}_{1} + {\vec{B}}_{2} = 0 \\ B_{1} \to due to left part \\ B_{2} \to due right part \end{aligned}$
$B_{1}, B_{2}$ at centroid are numerically equal & opposite to each other. (i,e. one is into the paper and the other is towards the reader)

Question Bank Series

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Question Bank Series

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