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Moving Charges & Magnetism

153425 The magnetic field at a centre of solenoid is $B$ . The solenoid is cut into two equal parts. For the same current, what is $B$ at the centre of new solenoid?

1

B

2

2 B

3

4 B

4

\frac{B}{2}

Explanation:

A The magnitude of the field B inside a long solenoid carrying a current $I$ is
$B = μ_{0} nI$
Where, number of turns per unit length $(n) = \frac{N}{L}$
When, solenoid cut into two equal parts
$n^{'} = \frac{N / 2}{L / 2} = \frac{N}{L} = n$
Now magnetic field
$B^{'} = μ_{0} n^{'} I$
$B^{'} = μ_{0} nI$
$B^{'} = B$

TS EAMCET 31.07.2022

Moving Charges & Magnetism

153426 A solenoid having 2000 turns per meter has a core of area $4 {cm}^{2}$ and relative permeability 220. It carries a current of $5 A$ . The which statement among the following is true?

1 Magnetic intensity is

10^{4} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 876 A.m

2 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 200 A.m

3 Amgnetic intensity is

10^{4}

A.m

^{- 1}

. Magnetisation of core is

2.19 \times 10^{6} A^{- m^{- 1}}

and Pole strength is 300 A.m

4 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

44 π \times 10^{- 2} T

and Magnetisation of core is

2.19 \times 10^{6} A \cdot m^{- 1}

Explanation:

A Given that, $n = 2000$ turn/meter, $I = 5 A, μ_{r} =$ $220, A = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
(i) Magnetic intensity $(H) = nI$
$H = 2000 \times 5 = 10000$
$H = 10^{4} A / m$
(ii) Magnetic field (B) $= μ H = μ_{0} μ_{r} H$
$B = 4 π \times 10^{- 7} \times 220 \times 10^{4}$
$= 88 π \times 10^{- 2} Tesla$
(iii) We know that, $B = B_{0} (H + M)$
$88 π \times 10^{- 2} = (10^{4} + M)$
$M = 2.19 \times 10^{6} A / m (M = I)$
(iv) Pole strength $(m) = I$ . A
$m = 2.19 \times 10^{6} \times 4 \times 10^{- 4}$
$m = 876 A / m$

AP EAMCET-06.09.2021

Moving Charges & Magnetism

153427 In a current carrying long solenoid, the field produced does not depend upon

1 Number of turns per unit length

2 Current flowing

3 Radius of the solenoid

4 All of the above

Explanation:

C The magnetic field at the center of solenoid is
$B = μ_{o} NI$
Where, $N \to$ no of turns per unit length
$I \to current$
$μ_{o} \to 4 π \times 10^{- 7} H / m$
So, magnetic field does not depend on the radius of the solenoid.

CG PET-2021

Moving Charges & Magnetism

153429 An ideal solenoid has $1000$ turns per meter and $8 cm$ radius. The current in it varies at a uniform rate of $0.02 A s^{- 1}$ . A circular coil of radius $2 cm$ is placed inside the solenoid such that its axis coincides with that of the solenoid. Find the induced electric field at a point on the circumference of the coil and that at a point outside the solenoid at a distance of $12.8 cm$ from its axis.

1

8 π \times 10^{- 8} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

2

7 π \times 10^{- 7} V m^{- 1}

and

5 π \times 10^{- 7} V m^{- 1}

3

6 π \times 10^{- 7} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

4

8 π \times 10^{- 8} V m^{- 1}

and

28 π \times 10^{- 7} V m^{- 1}

Explanation:

A Given that, $N = 1000$ turn/meter, $R = 8 \times 10^{- 2}$ $m, \frac{dI}{dt} = 0.02 A / \sec, r = 2 \times 10^{- 2} m$
We know that,
Total magnetic flux $(ϕ) = BA = μ_{0} n$ I A
$(i) \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$\frac{d ϕ}{dt} = 4 π \times 10^{- 7} \times 1000 \times π {(2 \times 10^{- 2})}^{2} \times 0.02$
$\frac{d ϕ}{dt} = 32 π^{2} \times 10^{- 10}$
Therefore, induced electric field
$\int E . d l = \frac{d ϕ}{dt}$
$E = \frac{32 π^{2} \times 10^{- 10}}{2 π r} = \frac{32 π^{2} \times 10^{- 10}}{2 π \times 2 \times 10^{- 2}}$
$E = 8 π \times 10^{- 8} V / m$
(ii) Induced electric field at distance $12.8 \times 10^{- 2} m$ then
$\int E . d l = \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$= 4 π \times 10^{- 7} \times 1000 \times π \times {(8 \times 10^{- 2})}^{2} \times 0.02$
$∵ E = \frac{4 π \times 10^{- 7} \times 1000 \times π \times 64 \times 10^{- 4} \times 0.02}{2 π \times 12.8 \times 10^{- 2}}$
$E = 2 π \times 10^{- 7} V / m$

AP EAMCET-03.09.2021

Moving Charges & Magnetism

153430 Two long parallel wires are separated by a distance of $2.50 cm$ . The force per unit length that each wire exerts on the other is $4 \times 10^{-}$ $^{5} N / m$ , and the wires repel each other. The current in one wire is $0.5 A$ . What is the current in the second wire?
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI Unit)

1

12 A

2

8 A

3

6 A

4

10 A

Explanation:

D Given, force per unit length $(F / L) = 4 \times 10^{- 5}$
$N / m$ , current in one wire $(i_{1}) = 0.5 A$ , wire separated $(r) =$ $2.50 cm = 2.5 \times 10^{- 2} m$
We know that,
$\frac{F}{L} = (\frac{μ_{o}}{2 π}) \frac{i_{1} i_{2}}{r}$
$4 \times 10^{- 5} = (\frac{4 π \times 10^{- 7}}{2 π}) \times (\frac{0.5 \times i_{2}}{2.5 \times 10^{- 2}})$
$i_{2} = \frac{4 \times 10^{- 5} \times 2.5 \times 10^{- 2}}{2 \times 10^{- 7} \times 0.5}$
$i_{2} = 10 A$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153425 The magnetic field at a centre of solenoid is $B$ . The solenoid is cut into two equal parts. For the same current, what is $B$ at the centre of new solenoid?

1

B

2

2 B

3

4 B

4

\frac{B}{2}

Explanation:

A The magnitude of the field B inside a long solenoid carrying a current $I$ is
$B = μ_{0} nI$
Where, number of turns per unit length $(n) = \frac{N}{L}$
When, solenoid cut into two equal parts
$n^{'} = \frac{N / 2}{L / 2} = \frac{N}{L} = n$
Now magnetic field
$B^{'} = μ_{0} n^{'} I$
$B^{'} = μ_{0} nI$
$B^{'} = B$

TS EAMCET 31.07.2022

Moving Charges & Magnetism

153426 A solenoid having 2000 turns per meter has a core of area $4 {cm}^{2}$ and relative permeability 220. It carries a current of $5 A$ . The which statement among the following is true?

1 Magnetic intensity is

10^{4} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 876 A.m

2 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 200 A.m

3 Amgnetic intensity is

10^{4}

A.m

^{- 1}

. Magnetisation of core is

2.19 \times 10^{6} A^{- m^{- 1}}

and Pole strength is 300 A.m

4 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

44 π \times 10^{- 2} T

and Magnetisation of core is

2.19 \times 10^{6} A \cdot m^{- 1}

Explanation:

A Given that, $n = 2000$ turn/meter, $I = 5 A, μ_{r} =$ $220, A = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
(i) Magnetic intensity $(H) = nI$
$H = 2000 \times 5 = 10000$
$H = 10^{4} A / m$
(ii) Magnetic field (B) $= μ H = μ_{0} μ_{r} H$
$B = 4 π \times 10^{- 7} \times 220 \times 10^{4}$
$= 88 π \times 10^{- 2} Tesla$
(iii) We know that, $B = B_{0} (H + M)$
$88 π \times 10^{- 2} = (10^{4} + M)$
$M = 2.19 \times 10^{6} A / m (M = I)$
(iv) Pole strength $(m) = I$ . A
$m = 2.19 \times 10^{6} \times 4 \times 10^{- 4}$
$m = 876 A / m$

AP EAMCET-06.09.2021

Moving Charges & Magnetism

153427 In a current carrying long solenoid, the field produced does not depend upon

1 Number of turns per unit length

2 Current flowing

3 Radius of the solenoid

4 All of the above

Explanation:

C The magnetic field at the center of solenoid is
$B = μ_{o} NI$
Where, $N \to$ no of turns per unit length
$I \to current$
$μ_{o} \to 4 π \times 10^{- 7} H / m$
So, magnetic field does not depend on the radius of the solenoid.

CG PET-2021

Moving Charges & Magnetism

153429 An ideal solenoid has $1000$ turns per meter and $8 cm$ radius. The current in it varies at a uniform rate of $0.02 A s^{- 1}$ . A circular coil of radius $2 cm$ is placed inside the solenoid such that its axis coincides with that of the solenoid. Find the induced electric field at a point on the circumference of the coil and that at a point outside the solenoid at a distance of $12.8 cm$ from its axis.

1

8 π \times 10^{- 8} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

2

7 π \times 10^{- 7} V m^{- 1}

and

5 π \times 10^{- 7} V m^{- 1}

3

6 π \times 10^{- 7} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

4

8 π \times 10^{- 8} V m^{- 1}

and

28 π \times 10^{- 7} V m^{- 1}

Explanation:

A Given that, $N = 1000$ turn/meter, $R = 8 \times 10^{- 2}$ $m, \frac{dI}{dt} = 0.02 A / \sec, r = 2 \times 10^{- 2} m$
We know that,
Total magnetic flux $(ϕ) = BA = μ_{0} n$ I A
$(i) \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$\frac{d ϕ}{dt} = 4 π \times 10^{- 7} \times 1000 \times π {(2 \times 10^{- 2})}^{2} \times 0.02$
$\frac{d ϕ}{dt} = 32 π^{2} \times 10^{- 10}$
Therefore, induced electric field
$\int E . d l = \frac{d ϕ}{dt}$
$E = \frac{32 π^{2} \times 10^{- 10}}{2 π r} = \frac{32 π^{2} \times 10^{- 10}}{2 π \times 2 \times 10^{- 2}}$
$E = 8 π \times 10^{- 8} V / m$
(ii) Induced electric field at distance $12.8 \times 10^{- 2} m$ then
$\int E . d l = \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$= 4 π \times 10^{- 7} \times 1000 \times π \times {(8 \times 10^{- 2})}^{2} \times 0.02$
$∵ E = \frac{4 π \times 10^{- 7} \times 1000 \times π \times 64 \times 10^{- 4} \times 0.02}{2 π \times 12.8 \times 10^{- 2}}$
$E = 2 π \times 10^{- 7} V / m$

AP EAMCET-03.09.2021

Moving Charges & Magnetism

153430 Two long parallel wires are separated by a distance of $2.50 cm$ . The force per unit length that each wire exerts on the other is $4 \times 10^{-}$ $^{5} N / m$ , and the wires repel each other. The current in one wire is $0.5 A$ . What is the current in the second wire?
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI Unit)

1

12 A

2

8 A

3

6 A

4

10 A

Explanation:

D Given, force per unit length $(F / L) = 4 \times 10^{- 5}$
$N / m$ , current in one wire $(i_{1}) = 0.5 A$ , wire separated $(r) =$ $2.50 cm = 2.5 \times 10^{- 2} m$
We know that,
$\frac{F}{L} = (\frac{μ_{o}}{2 π}) \frac{i_{1} i_{2}}{r}$
$4 \times 10^{- 5} = (\frac{4 π \times 10^{- 7}}{2 π}) \times (\frac{0.5 \times i_{2}}{2.5 \times 10^{- 2}})$
$i_{2} = \frac{4 \times 10^{- 5} \times 2.5 \times 10^{- 2}}{2 \times 10^{- 7} \times 0.5}$
$i_{2} = 10 A$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153425 The magnetic field at a centre of solenoid is $B$ . The solenoid is cut into two equal parts. For the same current, what is $B$ at the centre of new solenoid?

1

B

2

2 B

3

4 B

4

\frac{B}{2}

Explanation:

A The magnitude of the field B inside a long solenoid carrying a current $I$ is
$B = μ_{0} nI$
Where, number of turns per unit length $(n) = \frac{N}{L}$
When, solenoid cut into two equal parts
$n^{'} = \frac{N / 2}{L / 2} = \frac{N}{L} = n$
Now magnetic field
$B^{'} = μ_{0} n^{'} I$
$B^{'} = μ_{0} nI$
$B^{'} = B$

TS EAMCET 31.07.2022

Moving Charges & Magnetism

153426 A solenoid having 2000 turns per meter has a core of area $4 {cm}^{2}$ and relative permeability 220. It carries a current of $5 A$ . The which statement among the following is true?

1 Magnetic intensity is

10^{4} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 876 A.m

2 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 200 A.m

3 Amgnetic intensity is

10^{4}

A.m

^{- 1}

. Magnetisation of core is

2.19 \times 10^{6} A^{- m^{- 1}}

and Pole strength is 300 A.m

4 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

44 π \times 10^{- 2} T

and Magnetisation of core is

2.19 \times 10^{6} A \cdot m^{- 1}

Explanation:

A Given that, $n = 2000$ turn/meter, $I = 5 A, μ_{r} =$ $220, A = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
(i) Magnetic intensity $(H) = nI$
$H = 2000 \times 5 = 10000$
$H = 10^{4} A / m$
(ii) Magnetic field (B) $= μ H = μ_{0} μ_{r} H$
$B = 4 π \times 10^{- 7} \times 220 \times 10^{4}$
$= 88 π \times 10^{- 2} Tesla$
(iii) We know that, $B = B_{0} (H + M)$
$88 π \times 10^{- 2} = (10^{4} + M)$
$M = 2.19 \times 10^{6} A / m (M = I)$
(iv) Pole strength $(m) = I$ . A
$m = 2.19 \times 10^{6} \times 4 \times 10^{- 4}$
$m = 876 A / m$

AP EAMCET-06.09.2021

Moving Charges & Magnetism

153427 In a current carrying long solenoid, the field produced does not depend upon

1 Number of turns per unit length

2 Current flowing

3 Radius of the solenoid

4 All of the above

Explanation:

C The magnetic field at the center of solenoid is
$B = μ_{o} NI$
Where, $N \to$ no of turns per unit length
$I \to current$
$μ_{o} \to 4 π \times 10^{- 7} H / m$
So, magnetic field does not depend on the radius of the solenoid.

CG PET-2021

Moving Charges & Magnetism

153429 An ideal solenoid has $1000$ turns per meter and $8 cm$ radius. The current in it varies at a uniform rate of $0.02 A s^{- 1}$ . A circular coil of radius $2 cm$ is placed inside the solenoid such that its axis coincides with that of the solenoid. Find the induced electric field at a point on the circumference of the coil and that at a point outside the solenoid at a distance of $12.8 cm$ from its axis.

1

8 π \times 10^{- 8} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

2

7 π \times 10^{- 7} V m^{- 1}

and

5 π \times 10^{- 7} V m^{- 1}

3

6 π \times 10^{- 7} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

4

8 π \times 10^{- 8} V m^{- 1}

and

28 π \times 10^{- 7} V m^{- 1}

Explanation:

A Given that, $N = 1000$ turn/meter, $R = 8 \times 10^{- 2}$ $m, \frac{dI}{dt} = 0.02 A / \sec, r = 2 \times 10^{- 2} m$
We know that,
Total magnetic flux $(ϕ) = BA = μ_{0} n$ I A
$(i) \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$\frac{d ϕ}{dt} = 4 π \times 10^{- 7} \times 1000 \times π {(2 \times 10^{- 2})}^{2} \times 0.02$
$\frac{d ϕ}{dt} = 32 π^{2} \times 10^{- 10}$
Therefore, induced electric field
$\int E . d l = \frac{d ϕ}{dt}$
$E = \frac{32 π^{2} \times 10^{- 10}}{2 π r} = \frac{32 π^{2} \times 10^{- 10}}{2 π \times 2 \times 10^{- 2}}$
$E = 8 π \times 10^{- 8} V / m$
(ii) Induced electric field at distance $12.8 \times 10^{- 2} m$ then
$\int E . d l = \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$= 4 π \times 10^{- 7} \times 1000 \times π \times {(8 \times 10^{- 2})}^{2} \times 0.02$
$∵ E = \frac{4 π \times 10^{- 7} \times 1000 \times π \times 64 \times 10^{- 4} \times 0.02}{2 π \times 12.8 \times 10^{- 2}}$
$E = 2 π \times 10^{- 7} V / m$

AP EAMCET-03.09.2021

Moving Charges & Magnetism

153430 Two long parallel wires are separated by a distance of $2.50 cm$ . The force per unit length that each wire exerts on the other is $4 \times 10^{-}$ $^{5} N / m$ , and the wires repel each other. The current in one wire is $0.5 A$ . What is the current in the second wire?
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI Unit)

1

12 A

2

8 A

3

6 A

4

10 A

Explanation:

D Given, force per unit length $(F / L) = 4 \times 10^{- 5}$
$N / m$ , current in one wire $(i_{1}) = 0.5 A$ , wire separated $(r) =$ $2.50 cm = 2.5 \times 10^{- 2} m$
We know that,
$\frac{F}{L} = (\frac{μ_{o}}{2 π}) \frac{i_{1} i_{2}}{r}$
$4 \times 10^{- 5} = (\frac{4 π \times 10^{- 7}}{2 π}) \times (\frac{0.5 \times i_{2}}{2.5 \times 10^{- 2}})$
$i_{2} = \frac{4 \times 10^{- 5} \times 2.5 \times 10^{- 2}}{2 \times 10^{- 7} \times 0.5}$
$i_{2} = 10 A$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153425 The magnetic field at a centre of solenoid is $B$ . The solenoid is cut into two equal parts. For the same current, what is $B$ at the centre of new solenoid?

1

B

2

2 B

3

4 B

4

\frac{B}{2}

Explanation:

A The magnitude of the field B inside a long solenoid carrying a current $I$ is
$B = μ_{0} nI$
Where, number of turns per unit length $(n) = \frac{N}{L}$
When, solenoid cut into two equal parts
$n^{'} = \frac{N / 2}{L / 2} = \frac{N}{L} = n$
Now magnetic field
$B^{'} = μ_{0} n^{'} I$
$B^{'} = μ_{0} nI$
$B^{'} = B$

TS EAMCET 31.07.2022

Moving Charges & Magnetism

153426 A solenoid having 2000 turns per meter has a core of area $4 {cm}^{2}$ and relative permeability 220. It carries a current of $5 A$ . The which statement among the following is true?

1 Magnetic intensity is

10^{4} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 876 A.m

2 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 200 A.m

3 Amgnetic intensity is

10^{4}

A.m

^{- 1}

. Magnetisation of core is

2.19 \times 10^{6} A^{- m^{- 1}}

and Pole strength is 300 A.m

4 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

44 π \times 10^{- 2} T

and Magnetisation of core is

2.19 \times 10^{6} A \cdot m^{- 1}

Explanation:

A Given that, $n = 2000$ turn/meter, $I = 5 A, μ_{r} =$ $220, A = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
(i) Magnetic intensity $(H) = nI$
$H = 2000 \times 5 = 10000$
$H = 10^{4} A / m$
(ii) Magnetic field (B) $= μ H = μ_{0} μ_{r} H$
$B = 4 π \times 10^{- 7} \times 220 \times 10^{4}$
$= 88 π \times 10^{- 2} Tesla$
(iii) We know that, $B = B_{0} (H + M)$
$88 π \times 10^{- 2} = (10^{4} + M)$
$M = 2.19 \times 10^{6} A / m (M = I)$
(iv) Pole strength $(m) = I$ . A
$m = 2.19 \times 10^{6} \times 4 \times 10^{- 4}$
$m = 876 A / m$

AP EAMCET-06.09.2021

Moving Charges & Magnetism

153427 In a current carrying long solenoid, the field produced does not depend upon

1 Number of turns per unit length

2 Current flowing

3 Radius of the solenoid

4 All of the above

Explanation:

C The magnetic field at the center of solenoid is
$B = μ_{o} NI$
Where, $N \to$ no of turns per unit length
$I \to current$
$μ_{o} \to 4 π \times 10^{- 7} H / m$
So, magnetic field does not depend on the radius of the solenoid.

CG PET-2021

Moving Charges & Magnetism

153429 An ideal solenoid has $1000$ turns per meter and $8 cm$ radius. The current in it varies at a uniform rate of $0.02 A s^{- 1}$ . A circular coil of radius $2 cm$ is placed inside the solenoid such that its axis coincides with that of the solenoid. Find the induced electric field at a point on the circumference of the coil and that at a point outside the solenoid at a distance of $12.8 cm$ from its axis.

1

8 π \times 10^{- 8} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

2

7 π \times 10^{- 7} V m^{- 1}

and

5 π \times 10^{- 7} V m^{- 1}

3

6 π \times 10^{- 7} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

4

8 π \times 10^{- 8} V m^{- 1}

and

28 π \times 10^{- 7} V m^{- 1}

Explanation:

A Given that, $N = 1000$ turn/meter, $R = 8 \times 10^{- 2}$ $m, \frac{dI}{dt} = 0.02 A / \sec, r = 2 \times 10^{- 2} m$
We know that,
Total magnetic flux $(ϕ) = BA = μ_{0} n$ I A
$(i) \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$\frac{d ϕ}{dt} = 4 π \times 10^{- 7} \times 1000 \times π {(2 \times 10^{- 2})}^{2} \times 0.02$
$\frac{d ϕ}{dt} = 32 π^{2} \times 10^{- 10}$
Therefore, induced electric field
$\int E . d l = \frac{d ϕ}{dt}$
$E = \frac{32 π^{2} \times 10^{- 10}}{2 π r} = \frac{32 π^{2} \times 10^{- 10}}{2 π \times 2 \times 10^{- 2}}$
$E = 8 π \times 10^{- 8} V / m$
(ii) Induced electric field at distance $12.8 \times 10^{- 2} m$ then
$\int E . d l = \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$= 4 π \times 10^{- 7} \times 1000 \times π \times {(8 \times 10^{- 2})}^{2} \times 0.02$
$∵ E = \frac{4 π \times 10^{- 7} \times 1000 \times π \times 64 \times 10^{- 4} \times 0.02}{2 π \times 12.8 \times 10^{- 2}}$
$E = 2 π \times 10^{- 7} V / m$

AP EAMCET-03.09.2021

Moving Charges & Magnetism

153430 Two long parallel wires are separated by a distance of $2.50 cm$ . The force per unit length that each wire exerts on the other is $4 \times 10^{-}$ $^{5} N / m$ , and the wires repel each other. The current in one wire is $0.5 A$ . What is the current in the second wire?
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI Unit)

1

12 A

2

8 A

3

6 A

4

10 A

Explanation:

D Given, force per unit length $(F / L) = 4 \times 10^{- 5}$
$N / m$ , current in one wire $(i_{1}) = 0.5 A$ , wire separated $(r) =$ $2.50 cm = 2.5 \times 10^{- 2} m$
We know that,
$\frac{F}{L} = (\frac{μ_{o}}{2 π}) \frac{i_{1} i_{2}}{r}$
$4 \times 10^{- 5} = (\frac{4 π \times 10^{- 7}}{2 π}) \times (\frac{0.5 \times i_{2}}{2.5 \times 10^{- 2}})$
$i_{2} = \frac{4 \times 10^{- 5} \times 2.5 \times 10^{- 2}}{2 \times 10^{- 7} \times 0.5}$
$i_{2} = 10 A$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153425 The magnetic field at a centre of solenoid is $B$ . The solenoid is cut into two equal parts. For the same current, what is $B$ at the centre of new solenoid?

1

B

2

2 B

3

4 B

4

\frac{B}{2}

Explanation:

A The magnitude of the field B inside a long solenoid carrying a current $I$ is
$B = μ_{0} nI$
Where, number of turns per unit length $(n) = \frac{N}{L}$
When, solenoid cut into two equal parts
$n^{'} = \frac{N / 2}{L / 2} = \frac{N}{L} = n$
Now magnetic field
$B^{'} = μ_{0} n^{'} I$
$B^{'} = μ_{0} nI$
$B^{'} = B$

TS EAMCET 31.07.2022

Moving Charges & Magnetism

153426 A solenoid having 2000 turns per meter has a core of area $4 {cm}^{2}$ and relative permeability 220. It carries a current of $5 A$ . The which statement among the following is true?

1 Magnetic intensity is

10^{4} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 876 A.m

2 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

88 π \times 10^{- 2} T

and Pole strength is 200 A.m

3 Amgnetic intensity is

10^{4}

A.m

^{- 1}

. Magnetisation of core is

2.19 \times 10^{6} A^{- m^{- 1}}

and Pole strength is 300 A.m

4 Magnetic intensity is

10^{3} A \cdot m^{- 1}

. Magnetic field is

44 π \times 10^{- 2} T

and Magnetisation of core is

2.19 \times 10^{6} A \cdot m^{- 1}

Explanation:

A Given that, $n = 2000$ turn/meter, $I = 5 A, μ_{r} =$ $220, A = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
(i) Magnetic intensity $(H) = nI$
$H = 2000 \times 5 = 10000$
$H = 10^{4} A / m$
(ii) Magnetic field (B) $= μ H = μ_{0} μ_{r} H$
$B = 4 π \times 10^{- 7} \times 220 \times 10^{4}$
$= 88 π \times 10^{- 2} Tesla$
(iii) We know that, $B = B_{0} (H + M)$
$88 π \times 10^{- 2} = (10^{4} + M)$
$M = 2.19 \times 10^{6} A / m (M = I)$
(iv) Pole strength $(m) = I$ . A
$m = 2.19 \times 10^{6} \times 4 \times 10^{- 4}$
$m = 876 A / m$

AP EAMCET-06.09.2021

Moving Charges & Magnetism

153427 In a current carrying long solenoid, the field produced does not depend upon

1 Number of turns per unit length

2 Current flowing

3 Radius of the solenoid

4 All of the above

Explanation:

C The magnetic field at the center of solenoid is
$B = μ_{o} NI$
Where, $N \to$ no of turns per unit length
$I \to current$
$μ_{o} \to 4 π \times 10^{- 7} H / m$
So, magnetic field does not depend on the radius of the solenoid.

CG PET-2021

Moving Charges & Magnetism

153429 An ideal solenoid has $1000$ turns per meter and $8 cm$ radius. The current in it varies at a uniform rate of $0.02 A s^{- 1}$ . A circular coil of radius $2 cm$ is placed inside the solenoid such that its axis coincides with that of the solenoid. Find the induced electric field at a point on the circumference of the coil and that at a point outside the solenoid at a distance of $12.8 cm$ from its axis.

1

8 π \times 10^{- 8} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

2

7 π \times 10^{- 7} V m^{- 1}

and

5 π \times 10^{- 7} V m^{- 1}

3

6 π \times 10^{- 7} V m^{- 1}

and

2 π \times 10^{- 7} V m^{- 1}

4

8 π \times 10^{- 8} V m^{- 1}

and

28 π \times 10^{- 7} V m^{- 1}

Explanation:

A Given that, $N = 1000$ turn/meter, $R = 8 \times 10^{- 2}$ $m, \frac{dI}{dt} = 0.02 A / \sec, r = 2 \times 10^{- 2} m$
We know that,
Total magnetic flux $(ϕ) = BA = μ_{0} n$ I A
$(i) \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$\frac{d ϕ}{dt} = 4 π \times 10^{- 7} \times 1000 \times π {(2 \times 10^{- 2})}^{2} \times 0.02$
$\frac{d ϕ}{dt} = 32 π^{2} \times 10^{- 10}$
Therefore, induced electric field
$\int E . d l = \frac{d ϕ}{dt}$
$E = \frac{32 π^{2} \times 10^{- 10}}{2 π r} = \frac{32 π^{2} \times 10^{- 10}}{2 π \times 2 \times 10^{- 2}}$
$E = 8 π \times 10^{- 8} V / m$
(ii) Induced electric field at distance $12.8 \times 10^{- 2} m$ then
$\int E . d l = \frac{d ϕ}{dt} = μ_{0} nA \frac{dI}{dt}$
$= 4 π \times 10^{- 7} \times 1000 \times π \times {(8 \times 10^{- 2})}^{2} \times 0.02$
$∵ E = \frac{4 π \times 10^{- 7} \times 1000 \times π \times 64 \times 10^{- 4} \times 0.02}{2 π \times 12.8 \times 10^{- 2}}$
$E = 2 π \times 10^{- 7} V / m$

AP EAMCET-03.09.2021

Moving Charges & Magnetism

153430 Two long parallel wires are separated by a distance of $2.50 cm$ . The force per unit length that each wire exerts on the other is $4 \times 10^{-}$ $^{5} N / m$ , and the wires repel each other. The current in one wire is $0.5 A$ . What is the current in the second wire?
(Take $μ_{0} = 4 π \times 10^{- 7}$ SI Unit)

1

12 A

2

8 A

3

6 A

4

10 A

Explanation:

D Given, force per unit length $(F / L) = 4 \times 10^{- 5}$
$N / m$ , current in one wire $(i_{1}) = 0.5 A$ , wire separated $(r) =$ $2.50 cm = 2.5 \times 10^{- 2} m$
We know that,
$\frac{F}{L} = (\frac{μ_{o}}{2 π}) \frac{i_{1} i_{2}}{r}$
$4 \times 10^{- 5} = (\frac{4 π \times 10^{- 7}}{2 π}) \times (\frac{0.5 \times i_{2}}{2.5 \times 10^{- 2}})$
$i_{2} = \frac{4 \times 10^{- 5} \times 2.5 \times 10^{- 2}}{2 \times 10^{- 7} \times 0.5}$
$i_{2} = 10 A$

TS EAMCET 05.08.2021