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PHXII04:MOVING CHARGES AND MAGNETISM

362566 Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $21$ and $41$ respectively. The resultant magnetic field induction at the centre is found to be $N$ times $\frac{μ_{0} I}{R}$ . Find the value of $N$ is (Take $\sqrt{5} = 2.24$ )

1 1.15

2 3.58

3 5.27

4 2.24

Explanation:

supporting img
Magnetic field induction due to vertical loop at centre $O$ is $B_{1} = \frac{μ_{0} 2 I}{2 R} \hat{i}$
Magnetic field due to horizontal loop at centre $O$ is $B_{2} = \frac{μ_{0} 4 I}{2 R} \hat{j}$
Now,
$B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(\frac{μ_{0} 2 I}{2 R})}^{2} + {(\frac{μ_{0} 4 I}{2 R})}^{2}}$
$= \frac{μ_{0} I}{2 R} \sqrt{{(2)}^{2} + {(4)}^{2}}$
$= \frac{\sqrt{20}}{2} \frac{μ_{0} I}{R}$
$B_{n e t} = \frac{2 \sqrt{5}}{2} = 2.24 \frac{μ_{0} I}{R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362567 Two concentric circular loops of radii $r_{1}$ and $r_{2}$ carry clockwise and anticlockwise currents $i_{1}$ and $i_{2}$ . If the centre is a null point, $i_{1} / i_{2}$ must be equal to

1

r_{2} / r_{1}

2

r_{2}^{2} / r_{1}^{2}

3

r_{1}^{2} / r_{2}^{2}

4

r_{1} / r_{2}

Explanation:

Magnetic field due to both loops should be equal and opposite to each other
$\begin{aligned} B_{1} = - B_{2} \\ \frac{μ_{o} i_{1}}{2 π r_{1}} = \frac{μ_{o} i_{2}}{2 π r_{2}} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{r_{1}}{r_{2}} \end{aligned}$

PHXII04:MOVING CHARGES AND MAGNETISM

362568 A circular coil carrying a certain current produces a magnetic field $B_{0}$ at its centre. The coil is now rewound so as to have 4 turns and the same current is passed through it. The new magnetic field at the centre is:

1

\frac{B_{0}}{16}

2

16 B_{0}

3

\frac{B_{0}}{4}

4

4 B_{0}

Explanation:

The magnetic field produced at the centre of the circular coil carrying current $I$ is given by
$B = \frac{μ_{0} N I}{2 r}$
For one turn, $N = 1$
$B_{0} = \frac{μ_{0} I}{2 r}$
As the coil is rewound
$\begin{matrix} r^{'} = \frac{r}{4} \cdot N^{'} = 4 \\ ∴ B^{'} = \frac{μ_{0} I \times 4}{2 \times (\frac{r}{4})} = \frac{16 μ_{0} I}{2 r} = 16 B_{0} \end{matrix}$

PHXII04:MOVING CHARGES AND MAGNETISM

362569 A coil having 9 turns carrying a current produces magnetic filed $B_{1}$ at the centre. Now the coil is rewounded into 3 turns carrying same current. Then the magnetic field at the centre $B_{2} =$

1

\frac{B_{1}}{9}

2

9 B_{1}

3

3 B_{1}

4

\frac{B_{1}}{3}

Explanation:

Magnetic at the centre of current carrying circular coil is given by
$B = \frac{μ_{0} N i}{R}$
supporting img
No. of turns $= N_{1} = 9$
$B_{1} = \frac{μ_{0} N_{1} i}{2 R} = \frac{9 μ_{0} i}{2 R}$

Similarly, $B_{2} = \frac{μ_{0} N_{2} i}{2 r} = \frac{μ_{0} i}{2 R}$
$\frac{B_{1}}{B_{2}} = \frac{\frac{9 μ_{0} i}{2 R}}{\frac{μ_{0} i}{2 R}} = 9 \Rightarrow B_{1} = 9 B_{2}$
$∴ B_{2} = \frac{B_{1}}{9}$

KCET - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362566 Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $21$ and $41$ respectively. The resultant magnetic field induction at the centre is found to be $N$ times $\frac{μ_{0} I}{R}$ . Find the value of $N$ is (Take $\sqrt{5} = 2.24$ )

1 1.15

2 3.58

3 5.27

4 2.24

Explanation:

supporting img
Magnetic field induction due to vertical loop at centre $O$ is $B_{1} = \frac{μ_{0} 2 I}{2 R} \hat{i}$
Magnetic field due to horizontal loop at centre $O$ is $B_{2} = \frac{μ_{0} 4 I}{2 R} \hat{j}$
Now,
$B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(\frac{μ_{0} 2 I}{2 R})}^{2} + {(\frac{μ_{0} 4 I}{2 R})}^{2}}$
$= \frac{μ_{0} I}{2 R} \sqrt{{(2)}^{2} + {(4)}^{2}}$
$= \frac{\sqrt{20}}{2} \frac{μ_{0} I}{R}$
$B_{n e t} = \frac{2 \sqrt{5}}{2} = 2.24 \frac{μ_{0} I}{R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362567 Two concentric circular loops of radii $r_{1}$ and $r_{2}$ carry clockwise and anticlockwise currents $i_{1}$ and $i_{2}$ . If the centre is a null point, $i_{1} / i_{2}$ must be equal to

1

r_{2} / r_{1}

2

r_{2}^{2} / r_{1}^{2}

3

r_{1}^{2} / r_{2}^{2}

4

r_{1} / r_{2}

Explanation:

Magnetic field due to both loops should be equal and opposite to each other
$\begin{aligned} B_{1} = - B_{2} \\ \frac{μ_{o} i_{1}}{2 π r_{1}} = \frac{μ_{o} i_{2}}{2 π r_{2}} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{r_{1}}{r_{2}} \end{aligned}$

PHXII04:MOVING CHARGES AND MAGNETISM

362568 A circular coil carrying a certain current produces a magnetic field $B_{0}$ at its centre. The coil is now rewound so as to have 4 turns and the same current is passed through it. The new magnetic field at the centre is:

1

\frac{B_{0}}{16}

2

16 B_{0}

3

\frac{B_{0}}{4}

4

4 B_{0}

Explanation:

The magnetic field produced at the centre of the circular coil carrying current $I$ is given by
$B = \frac{μ_{0} N I}{2 r}$
For one turn, $N = 1$
$B_{0} = \frac{μ_{0} I}{2 r}$
As the coil is rewound
$\begin{matrix} r^{'} = \frac{r}{4} \cdot N^{'} = 4 \\ ∴ B^{'} = \frac{μ_{0} I \times 4}{2 \times (\frac{r}{4})} = \frac{16 μ_{0} I}{2 r} = 16 B_{0} \end{matrix}$

PHXII04:MOVING CHARGES AND MAGNETISM

362569 A coil having 9 turns carrying a current produces magnetic filed $B_{1}$ at the centre. Now the coil is rewounded into 3 turns carrying same current. Then the magnetic field at the centre $B_{2} =$

1

\frac{B_{1}}{9}

2

9 B_{1}

3

3 B_{1}

4

\frac{B_{1}}{3}

Explanation:

Magnetic at the centre of current carrying circular coil is given by
$B = \frac{μ_{0} N i}{R}$
supporting img
No. of turns $= N_{1} = 9$
$B_{1} = \frac{μ_{0} N_{1} i}{2 R} = \frac{9 μ_{0} i}{2 R}$

Similarly, $B_{2} = \frac{μ_{0} N_{2} i}{2 r} = \frac{μ_{0} i}{2 R}$
$\frac{B_{1}}{B_{2}} = \frac{\frac{9 μ_{0} i}{2 R}}{\frac{μ_{0} i}{2 R}} = 9 \Rightarrow B_{1} = 9 B_{2}$
$∴ B_{2} = \frac{B_{1}}{9}$

KCET - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362566 Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $21$ and $41$ respectively. The resultant magnetic field induction at the centre is found to be $N$ times $\frac{μ_{0} I}{R}$ . Find the value of $N$ is (Take $\sqrt{5} = 2.24$ )

1 1.15

2 3.58

3 5.27

4 2.24

Explanation:

supporting img
Magnetic field induction due to vertical loop at centre $O$ is $B_{1} = \frac{μ_{0} 2 I}{2 R} \hat{i}$
Magnetic field due to horizontal loop at centre $O$ is $B_{2} = \frac{μ_{0} 4 I}{2 R} \hat{j}$
Now,
$B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(\frac{μ_{0} 2 I}{2 R})}^{2} + {(\frac{μ_{0} 4 I}{2 R})}^{2}}$
$= \frac{μ_{0} I}{2 R} \sqrt{{(2)}^{2} + {(4)}^{2}}$
$= \frac{\sqrt{20}}{2} \frac{μ_{0} I}{R}$
$B_{n e t} = \frac{2 \sqrt{5}}{2} = 2.24 \frac{μ_{0} I}{R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362567 Two concentric circular loops of radii $r_{1}$ and $r_{2}$ carry clockwise and anticlockwise currents $i_{1}$ and $i_{2}$ . If the centre is a null point, $i_{1} / i_{2}$ must be equal to

1

r_{2} / r_{1}

2

r_{2}^{2} / r_{1}^{2}

3

r_{1}^{2} / r_{2}^{2}

4

r_{1} / r_{2}

Explanation:

Magnetic field due to both loops should be equal and opposite to each other
$\begin{aligned} B_{1} = - B_{2} \\ \frac{μ_{o} i_{1}}{2 π r_{1}} = \frac{μ_{o} i_{2}}{2 π r_{2}} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{r_{1}}{r_{2}} \end{aligned}$

PHXII04:MOVING CHARGES AND MAGNETISM

362568 A circular coil carrying a certain current produces a magnetic field $B_{0}$ at its centre. The coil is now rewound so as to have 4 turns and the same current is passed through it. The new magnetic field at the centre is:

1

\frac{B_{0}}{16}

2

16 B_{0}

3

\frac{B_{0}}{4}

4

4 B_{0}

Explanation:

The magnetic field produced at the centre of the circular coil carrying current $I$ is given by
$B = \frac{μ_{0} N I}{2 r}$
For one turn, $N = 1$
$B_{0} = \frac{μ_{0} I}{2 r}$
As the coil is rewound
$\begin{matrix} r^{'} = \frac{r}{4} \cdot N^{'} = 4 \\ ∴ B^{'} = \frac{μ_{0} I \times 4}{2 \times (\frac{r}{4})} = \frac{16 μ_{0} I}{2 r} = 16 B_{0} \end{matrix}$

PHXII04:MOVING CHARGES AND MAGNETISM

362569 A coil having 9 turns carrying a current produces magnetic filed $B_{1}$ at the centre. Now the coil is rewounded into 3 turns carrying same current. Then the magnetic field at the centre $B_{2} =$

1

\frac{B_{1}}{9}

2

9 B_{1}

3

3 B_{1}

4

\frac{B_{1}}{3}

Explanation:

Magnetic at the centre of current carrying circular coil is given by
$B = \frac{μ_{0} N i}{R}$
supporting img
No. of turns $= N_{1} = 9$
$B_{1} = \frac{μ_{0} N_{1} i}{2 R} = \frac{9 μ_{0} i}{2 R}$

Similarly, $B_{2} = \frac{μ_{0} N_{2} i}{2 r} = \frac{μ_{0} i}{2 R}$
$\frac{B_{1}}{B_{2}} = \frac{\frac{9 μ_{0} i}{2 R}}{\frac{μ_{0} i}{2 R}} = 9 \Rightarrow B_{1} = 9 B_{2}$
$∴ B_{2} = \frac{B_{1}}{9}$

KCET - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362566 Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $21$ and $41$ respectively. The resultant magnetic field induction at the centre is found to be $N$ times $\frac{μ_{0} I}{R}$ . Find the value of $N$ is (Take $\sqrt{5} = 2.24$ )

1 1.15

2 3.58

3 5.27

4 2.24

Explanation:

supporting img
Magnetic field induction due to vertical loop at centre $O$ is $B_{1} = \frac{μ_{0} 2 I}{2 R} \hat{i}$
Magnetic field due to horizontal loop at centre $O$ is $B_{2} = \frac{μ_{0} 4 I}{2 R} \hat{j}$
Now,
$B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(\frac{μ_{0} 2 I}{2 R})}^{2} + {(\frac{μ_{0} 4 I}{2 R})}^{2}}$
$= \frac{μ_{0} I}{2 R} \sqrt{{(2)}^{2} + {(4)}^{2}}$
$= \frac{\sqrt{20}}{2} \frac{μ_{0} I}{R}$
$B_{n e t} = \frac{2 \sqrt{5}}{2} = 2.24 \frac{μ_{0} I}{R}$

PHXII04:MOVING CHARGES AND MAGNETISM

362567 Two concentric circular loops of radii $r_{1}$ and $r_{2}$ carry clockwise and anticlockwise currents $i_{1}$ and $i_{2}$ . If the centre is a null point, $i_{1} / i_{2}$ must be equal to

1

r_{2} / r_{1}

2

r_{2}^{2} / r_{1}^{2}

3

r_{1}^{2} / r_{2}^{2}

4

r_{1} / r_{2}

Explanation:

Magnetic field due to both loops should be equal and opposite to each other
$\begin{aligned} B_{1} = - B_{2} \\ \frac{μ_{o} i_{1}}{2 π r_{1}} = \frac{μ_{o} i_{2}}{2 π r_{2}} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{r_{1}}{r_{2}} \end{aligned}$

PHXII04:MOVING CHARGES AND MAGNETISM

362568 A circular coil carrying a certain current produces a magnetic field $B_{0}$ at its centre. The coil is now rewound so as to have 4 turns and the same current is passed through it. The new magnetic field at the centre is:

1

\frac{B_{0}}{16}

2

16 B_{0}

3

\frac{B_{0}}{4}

4

4 B_{0}

Explanation:

The magnetic field produced at the centre of the circular coil carrying current $I$ is given by
$B = \frac{μ_{0} N I}{2 r}$
For one turn, $N = 1$
$B_{0} = \frac{μ_{0} I}{2 r}$
As the coil is rewound
$\begin{matrix} r^{'} = \frac{r}{4} \cdot N^{'} = 4 \\ ∴ B^{'} = \frac{μ_{0} I \times 4}{2 \times (\frac{r}{4})} = \frac{16 μ_{0} I}{2 r} = 16 B_{0} \end{matrix}$

PHXII04:MOVING CHARGES AND MAGNETISM

362569 A coil having 9 turns carrying a current produces magnetic filed $B_{1}$ at the centre. Now the coil is rewounded into 3 turns carrying same current. Then the magnetic field at the centre $B_{2} =$

1

\frac{B_{1}}{9}

2

9 B_{1}

3

3 B_{1}

4

\frac{B_{1}}{3}

Explanation:

Magnetic at the centre of current carrying circular coil is given by
$B = \frac{μ_{0} N i}{R}$
supporting img
No. of turns $= N_{1} = 9$
$B_{1} = \frac{μ_{0} N_{1} i}{2 R} = \frac{9 μ_{0} i}{2 R}$

Similarly, $B_{2} = \frac{μ_{0} N_{2} i}{2 r} = \frac{μ_{0} i}{2 R}$
$\frac{B_{1}}{B_{2}} = \frac{\frac{9 μ_{0} i}{2 R}}{\frac{μ_{0} i}{2 R}} = 9 \Rightarrow B_{1} = 9 B_{2}$
$∴ B_{2} = \frac{B_{1}}{9}$

KCET - 2024