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Moving Charges & Magnetism

153410 The free space inside a current carrying toroid is filled with a material of susceptibility
$2 \times 10^{- 2}$ . The percentage increase in the value of magnetic field inside the toroid will be

1

2 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

A Given, Susceptibility of material $(χ) = 2 \times$ $10^{- 2}$
We know, magnetic field in toroid
$B = μ nI$
Then, magnetic field in toroid with suspectibility
$B^{'} = μ_{0} μ_{r} nI$
$B^{'} = μ_{0} (1 + χ) nI (∵ μ_{r} = 1 + χ)$
Percentage increase in the value of magnetic field
$% Increase = \frac{B^{'} - B}{B}$
$= \frac{(χ + 1) - 1}{1} \times 100$
$= χ \times 100 = 2 \times 10^{- 2} \times 100 = 2 %$

JEE Main-11.04.2023

Moving Charges & Magnetism

153411 A long solenoid is formed by winding 70 turns ${cm}^{- 1}$ . If $2.0 A$ current flows, then the magnetic field produced inside the solenoid is $(μ_{0} = 4 π \times 10^{- 7}) {TmA}^{- 1})$

1

88 \times 10^{- 4} T

2

1232 \times 10^{- 4} T

3

352 \times 10^{- 4} T

4

176 \times 10^{- 4} T

Explanation:

D Given,
Number of turns in solenoid $(n) = 70$ turns ${cm}^{- 1}$
Flow of current (i) $= 2 A$
We know that,
$B = μ_{0} ni$
$B = 4 π \times 10^{- 7} \times \frac{70}{10^{- 2}} \times 2$
$B = 4 π \times 10^{- 5} \times 70 \times 2$
$B = 17.58 \times 10^{- 3} T$
$B = 176 \times 10^{- 4} T$
Hence, magnetic field produce inside the solenoid $(B) =$ $176 \times 10^{- 4} T$ .

JEE Main-24.01.2023

Moving Charges & Magnetism

153412 A long conducting wire having a current $i$ flowing it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

1

n : N

2

N : n

3

N^{2} : n^{2}

4

n^{2} : N^{2}

Explanation:

C Magnetic field when circular coil bent $N$ turns,
$B_{1} = \frac{μ_{0} i \times N}{2 R}$
Magnetic field when circular coil bent $n$ turns,
$B_{2} = \frac{μ_{0} i \times n}{2 R^{'}}$
But, length of the wire is same,
Then, $(2 π R) N = (2 π R^{'}) n$
So, $\frac{R^{'}}{R} = \frac{N}{n}$
From equation (i) and (ii),
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{R^{'}}{R})$
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{N}{n})$
$\frac{B_{1}}{B_{2}} = {(\frac{N}{n})}^{2}$

JEE Main-31.01.2023

Moving Charges & Magnetism

153413 A solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2 m$ long and $0.2 m$ in diameter. The magnetic intensity at the centre of the solenoid when a current of $2 A$ flows through it is,

1

2.4 \times 10^{3} {Am}^{- 1}

2

1.2 \times 10^{3} {Am}^{- 1}

3

1 {Am}^{- 1}

4

2.4 \times 10^{3} {Am}^{- 1}

Explanation:

B Given, no. of turn in solenoid (n) $= 1200$
Length of glass tube $(l) = 2 m$
Diameter $(d) = 0.2 m$
Flow of current $(i) = 2 A$
Number of turns per unit length $(n) = \frac{1200}{2} = 600$
We know that,
Magnetic intensity $= ni = 600 \times 2 = 1200 {Am}^{- 1}$ $= 1.2 \times 10^{3} {Am}^{- 1}$

JEE Main-25.01.2023

Moving Charges & Magnetism

153414 The magnitude of magnetic induction at mid point $O$ due to current arrangement as shown in Fig will be

1 0

2

\frac{μ_{0} l}{π a}

3

\frac{μ_{0} l}{2 π a}

4

\frac{μ_{0} l}{4 π a}

Explanation:

B From given figure,
Magnetic induction at point $O -$
$B_{O} = B_{AB} + B_{BC} + B_{CD} + B_{ET}$
$B_{O} = 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} + 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} = \frac{μ_{0} l}{π a}$

JEE Main-29.01.2023

Moving Charges & Magnetism

153410 The free space inside a current carrying toroid is filled with a material of susceptibility
$2 \times 10^{- 2}$ . The percentage increase in the value of magnetic field inside the toroid will be

1

2 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

A Given, Susceptibility of material $(χ) = 2 \times$ $10^{- 2}$
We know, magnetic field in toroid
$B = μ nI$
Then, magnetic field in toroid with suspectibility
$B^{'} = μ_{0} μ_{r} nI$
$B^{'} = μ_{0} (1 + χ) nI (∵ μ_{r} = 1 + χ)$
Percentage increase in the value of magnetic field
$% Increase = \frac{B^{'} - B}{B}$
$= \frac{(χ + 1) - 1}{1} \times 100$
$= χ \times 100 = 2 \times 10^{- 2} \times 100 = 2 %$

JEE Main-11.04.2023

Moving Charges & Magnetism

153411 A long solenoid is formed by winding 70 turns ${cm}^{- 1}$ . If $2.0 A$ current flows, then the magnetic field produced inside the solenoid is $(μ_{0} = 4 π \times 10^{- 7}) {TmA}^{- 1})$

1

88 \times 10^{- 4} T

2

1232 \times 10^{- 4} T

3

352 \times 10^{- 4} T

4

176 \times 10^{- 4} T

Explanation:

D Given,
Number of turns in solenoid $(n) = 70$ turns ${cm}^{- 1}$
Flow of current (i) $= 2 A$
We know that,
$B = μ_{0} ni$
$B = 4 π \times 10^{- 7} \times \frac{70}{10^{- 2}} \times 2$
$B = 4 π \times 10^{- 5} \times 70 \times 2$
$B = 17.58 \times 10^{- 3} T$
$B = 176 \times 10^{- 4} T$
Hence, magnetic field produce inside the solenoid $(B) =$ $176 \times 10^{- 4} T$ .

JEE Main-24.01.2023

Moving Charges & Magnetism

153412 A long conducting wire having a current $i$ flowing it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

1

n : N

2

N : n

3

N^{2} : n^{2}

4

n^{2} : N^{2}

Explanation:

C Magnetic field when circular coil bent $N$ turns,
$B_{1} = \frac{μ_{0} i \times N}{2 R}$
Magnetic field when circular coil bent $n$ turns,
$B_{2} = \frac{μ_{0} i \times n}{2 R^{'}}$
But, length of the wire is same,
Then, $(2 π R) N = (2 π R^{'}) n$
So, $\frac{R^{'}}{R} = \frac{N}{n}$
From equation (i) and (ii),
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{R^{'}}{R})$
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{N}{n})$
$\frac{B_{1}}{B_{2}} = {(\frac{N}{n})}^{2}$

JEE Main-31.01.2023

Moving Charges & Magnetism

153413 A solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2 m$ long and $0.2 m$ in diameter. The magnetic intensity at the centre of the solenoid when a current of $2 A$ flows through it is,

1

2.4 \times 10^{3} {Am}^{- 1}

2

1.2 \times 10^{3} {Am}^{- 1}

3

1 {Am}^{- 1}

4

2.4 \times 10^{3} {Am}^{- 1}

Explanation:

B Given, no. of turn in solenoid (n) $= 1200$
Length of glass tube $(l) = 2 m$
Diameter $(d) = 0.2 m$
Flow of current $(i) = 2 A$
Number of turns per unit length $(n) = \frac{1200}{2} = 600$
We know that,
Magnetic intensity $= ni = 600 \times 2 = 1200 {Am}^{- 1}$ $= 1.2 \times 10^{3} {Am}^{- 1}$

JEE Main-25.01.2023

Moving Charges & Magnetism

153414 The magnitude of magnetic induction at mid point $O$ due to current arrangement as shown in Fig will be

1 0

2

\frac{μ_{0} l}{π a}

3

\frac{μ_{0} l}{2 π a}

4

\frac{μ_{0} l}{4 π a}

Explanation:

B From given figure,
Magnetic induction at point $O -$
$B_{O} = B_{AB} + B_{BC} + B_{CD} + B_{ET}$
$B_{O} = 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} + 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} = \frac{μ_{0} l}{π a}$

JEE Main-29.01.2023

Moving Charges & Magnetism

153410 The free space inside a current carrying toroid is filled with a material of susceptibility
$2 \times 10^{- 2}$ . The percentage increase in the value of magnetic field inside the toroid will be

1

2 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

A Given, Susceptibility of material $(χ) = 2 \times$ $10^{- 2}$
We know, magnetic field in toroid
$B = μ nI$
Then, magnetic field in toroid with suspectibility
$B^{'} = μ_{0} μ_{r} nI$
$B^{'} = μ_{0} (1 + χ) nI (∵ μ_{r} = 1 + χ)$
Percentage increase in the value of magnetic field
$% Increase = \frac{B^{'} - B}{B}$
$= \frac{(χ + 1) - 1}{1} \times 100$
$= χ \times 100 = 2 \times 10^{- 2} \times 100 = 2 %$

JEE Main-11.04.2023

Moving Charges & Magnetism

153411 A long solenoid is formed by winding 70 turns ${cm}^{- 1}$ . If $2.0 A$ current flows, then the magnetic field produced inside the solenoid is $(μ_{0} = 4 π \times 10^{- 7}) {TmA}^{- 1})$

1

88 \times 10^{- 4} T

2

1232 \times 10^{- 4} T

3

352 \times 10^{- 4} T

4

176 \times 10^{- 4} T

Explanation:

D Given,
Number of turns in solenoid $(n) = 70$ turns ${cm}^{- 1}$
Flow of current (i) $= 2 A$
We know that,
$B = μ_{0} ni$
$B = 4 π \times 10^{- 7} \times \frac{70}{10^{- 2}} \times 2$
$B = 4 π \times 10^{- 5} \times 70 \times 2$
$B = 17.58 \times 10^{- 3} T$
$B = 176 \times 10^{- 4} T$
Hence, magnetic field produce inside the solenoid $(B) =$ $176 \times 10^{- 4} T$ .

JEE Main-24.01.2023

Moving Charges & Magnetism

153412 A long conducting wire having a current $i$ flowing it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

1

n : N

2

N : n

3

N^{2} : n^{2}

4

n^{2} : N^{2}

Explanation:

C Magnetic field when circular coil bent $N$ turns,
$B_{1} = \frac{μ_{0} i \times N}{2 R}$
Magnetic field when circular coil bent $n$ turns,
$B_{2} = \frac{μ_{0} i \times n}{2 R^{'}}$
But, length of the wire is same,
Then, $(2 π R) N = (2 π R^{'}) n$
So, $\frac{R^{'}}{R} = \frac{N}{n}$
From equation (i) and (ii),
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{R^{'}}{R})$
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{N}{n})$
$\frac{B_{1}}{B_{2}} = {(\frac{N}{n})}^{2}$

JEE Main-31.01.2023

Moving Charges & Magnetism

153413 A solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2 m$ long and $0.2 m$ in diameter. The magnetic intensity at the centre of the solenoid when a current of $2 A$ flows through it is,

1

2.4 \times 10^{3} {Am}^{- 1}

2

1.2 \times 10^{3} {Am}^{- 1}

3

1 {Am}^{- 1}

4

2.4 \times 10^{3} {Am}^{- 1}

Explanation:

B Given, no. of turn in solenoid (n) $= 1200$
Length of glass tube $(l) = 2 m$
Diameter $(d) = 0.2 m$
Flow of current $(i) = 2 A$
Number of turns per unit length $(n) = \frac{1200}{2} = 600$
We know that,
Magnetic intensity $= ni = 600 \times 2 = 1200 {Am}^{- 1}$ $= 1.2 \times 10^{3} {Am}^{- 1}$

JEE Main-25.01.2023

Moving Charges & Magnetism

153414 The magnitude of magnetic induction at mid point $O$ due to current arrangement as shown in Fig will be

1 0

2

\frac{μ_{0} l}{π a}

3

\frac{μ_{0} l}{2 π a}

4

\frac{μ_{0} l}{4 π a}

Explanation:

B From given figure,
Magnetic induction at point $O -$
$B_{O} = B_{AB} + B_{BC} + B_{CD} + B_{ET}$
$B_{O} = 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} + 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} = \frac{μ_{0} l}{π a}$

JEE Main-29.01.2023

Moving Charges & Magnetism

153410 The free space inside a current carrying toroid is filled with a material of susceptibility
$2 \times 10^{- 2}$ . The percentage increase in the value of magnetic field inside the toroid will be

1

2 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

A Given, Susceptibility of material $(χ) = 2 \times$ $10^{- 2}$
We know, magnetic field in toroid
$B = μ nI$
Then, magnetic field in toroid with suspectibility
$B^{'} = μ_{0} μ_{r} nI$
$B^{'} = μ_{0} (1 + χ) nI (∵ μ_{r} = 1 + χ)$
Percentage increase in the value of magnetic field
$% Increase = \frac{B^{'} - B}{B}$
$= \frac{(χ + 1) - 1}{1} \times 100$
$= χ \times 100 = 2 \times 10^{- 2} \times 100 = 2 %$

JEE Main-11.04.2023

Moving Charges & Magnetism

153411 A long solenoid is formed by winding 70 turns ${cm}^{- 1}$ . If $2.0 A$ current flows, then the magnetic field produced inside the solenoid is $(μ_{0} = 4 π \times 10^{- 7}) {TmA}^{- 1})$

1

88 \times 10^{- 4} T

2

1232 \times 10^{- 4} T

3

352 \times 10^{- 4} T

4

176 \times 10^{- 4} T

Explanation:

D Given,
Number of turns in solenoid $(n) = 70$ turns ${cm}^{- 1}$
Flow of current (i) $= 2 A$
We know that,
$B = μ_{0} ni$
$B = 4 π \times 10^{- 7} \times \frac{70}{10^{- 2}} \times 2$
$B = 4 π \times 10^{- 5} \times 70 \times 2$
$B = 17.58 \times 10^{- 3} T$
$B = 176 \times 10^{- 4} T$
Hence, magnetic field produce inside the solenoid $(B) =$ $176 \times 10^{- 4} T$ .

JEE Main-24.01.2023

Moving Charges & Magnetism

153412 A long conducting wire having a current $i$ flowing it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

1

n : N

2

N : n

3

N^{2} : n^{2}

4

n^{2} : N^{2}

Explanation:

C Magnetic field when circular coil bent $N$ turns,
$B_{1} = \frac{μ_{0} i \times N}{2 R}$
Magnetic field when circular coil bent $n$ turns,
$B_{2} = \frac{μ_{0} i \times n}{2 R^{'}}$
But, length of the wire is same,
Then, $(2 π R) N = (2 π R^{'}) n$
So, $\frac{R^{'}}{R} = \frac{N}{n}$
From equation (i) and (ii),
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{R^{'}}{R})$
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{N}{n})$
$\frac{B_{1}}{B_{2}} = {(\frac{N}{n})}^{2}$

JEE Main-31.01.2023

Moving Charges & Magnetism

153413 A solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2 m$ long and $0.2 m$ in diameter. The magnetic intensity at the centre of the solenoid when a current of $2 A$ flows through it is,

1

2.4 \times 10^{3} {Am}^{- 1}

2

1.2 \times 10^{3} {Am}^{- 1}

3

1 {Am}^{- 1}

4

2.4 \times 10^{3} {Am}^{- 1}

Explanation:

B Given, no. of turn in solenoid (n) $= 1200$
Length of glass tube $(l) = 2 m$
Diameter $(d) = 0.2 m$
Flow of current $(i) = 2 A$
Number of turns per unit length $(n) = \frac{1200}{2} = 600$
We know that,
Magnetic intensity $= ni = 600 \times 2 = 1200 {Am}^{- 1}$ $= 1.2 \times 10^{3} {Am}^{- 1}$

JEE Main-25.01.2023

Moving Charges & Magnetism

153414 The magnitude of magnetic induction at mid point $O$ due to current arrangement as shown in Fig will be

1 0

2

\frac{μ_{0} l}{π a}

3

\frac{μ_{0} l}{2 π a}

4

\frac{μ_{0} l}{4 π a}

Explanation:

B From given figure,
Magnetic induction at point $O -$
$B_{O} = B_{AB} + B_{BC} + B_{CD} + B_{ET}$
$B_{O} = 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} + 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} = \frac{μ_{0} l}{π a}$

JEE Main-29.01.2023

Moving Charges & Magnetism

153410 The free space inside a current carrying toroid is filled with a material of susceptibility
$2 \times 10^{- 2}$ . The percentage increase in the value of magnetic field inside the toroid will be

1

2 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

A Given, Susceptibility of material $(χ) = 2 \times$ $10^{- 2}$
We know, magnetic field in toroid
$B = μ nI$
Then, magnetic field in toroid with suspectibility
$B^{'} = μ_{0} μ_{r} nI$
$B^{'} = μ_{0} (1 + χ) nI (∵ μ_{r} = 1 + χ)$
Percentage increase in the value of magnetic field
$% Increase = \frac{B^{'} - B}{B}$
$= \frac{(χ + 1) - 1}{1} \times 100$
$= χ \times 100 = 2 \times 10^{- 2} \times 100 = 2 %$

JEE Main-11.04.2023

Moving Charges & Magnetism

153411 A long solenoid is formed by winding 70 turns ${cm}^{- 1}$ . If $2.0 A$ current flows, then the magnetic field produced inside the solenoid is $(μ_{0} = 4 π \times 10^{- 7}) {TmA}^{- 1})$

1

88 \times 10^{- 4} T

2

1232 \times 10^{- 4} T

3

352 \times 10^{- 4} T

4

176 \times 10^{- 4} T

Explanation:

D Given,
Number of turns in solenoid $(n) = 70$ turns ${cm}^{- 1}$
Flow of current (i) $= 2 A$
We know that,
$B = μ_{0} ni$
$B = 4 π \times 10^{- 7} \times \frac{70}{10^{- 2}} \times 2$
$B = 4 π \times 10^{- 5} \times 70 \times 2$
$B = 17.58 \times 10^{- 3} T$
$B = 176 \times 10^{- 4} T$
Hence, magnetic field produce inside the solenoid $(B) =$ $176 \times 10^{- 4} T$ .

JEE Main-24.01.2023

Moving Charges & Magnetism

153412 A long conducting wire having a current $i$ flowing it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

1

n : N

2

N : n

3

N^{2} : n^{2}

4

n^{2} : N^{2}

Explanation:

C Magnetic field when circular coil bent $N$ turns,
$B_{1} = \frac{μ_{0} i \times N}{2 R}$
Magnetic field when circular coil bent $n$ turns,
$B_{2} = \frac{μ_{0} i \times n}{2 R^{'}}$
But, length of the wire is same,
Then, $(2 π R) N = (2 π R^{'}) n$
So, $\frac{R^{'}}{R} = \frac{N}{n}$
From equation (i) and (ii),
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{R^{'}}{R})$
$\frac{B_{1}}{B_{2}} = (\frac{N}{n}) (\frac{N}{n})$
$\frac{B_{1}}{B_{2}} = {(\frac{N}{n})}^{2}$

JEE Main-31.01.2023

Moving Charges & Magnetism

153413 A solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2 m$ long and $0.2 m$ in diameter. The magnetic intensity at the centre of the solenoid when a current of $2 A$ flows through it is,

1

2.4 \times 10^{3} {Am}^{- 1}

2

1.2 \times 10^{3} {Am}^{- 1}

3

1 {Am}^{- 1}

4

2.4 \times 10^{3} {Am}^{- 1}

Explanation:

B Given, no. of turn in solenoid (n) $= 1200$
Length of glass tube $(l) = 2 m$
Diameter $(d) = 0.2 m$
Flow of current $(i) = 2 A$
Number of turns per unit length $(n) = \frac{1200}{2} = 600$
We know that,
Magnetic intensity $= ni = 600 \times 2 = 1200 {Am}^{- 1}$ $= 1.2 \times 10^{3} {Am}^{- 1}$

JEE Main-25.01.2023

Moving Charges & Magnetism

153414 The magnitude of magnetic induction at mid point $O$ due to current arrangement as shown in Fig will be

1 0

2

\frac{μ_{0} l}{π a}

3

\frac{μ_{0} l}{2 π a}

4

\frac{μ_{0} l}{4 π a}

Explanation:

B From given figure,
Magnetic induction at point $O -$
$B_{O} = B_{AB} + B_{BC} + B_{CD} + B_{ET}$
$B_{O} = 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} + 0 + \frac{μ_{0} l}{4 π \frac{a}{2}} = \frac{μ_{0} l}{π a}$

JEE Main-29.01.2023