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Moving Charges & Magnetism

153166 A long cylindrical wire of radius $R$ carries a uniform current $I$ flowing through it. The variation of magnetic field with distance $r$ from the axis of the wire is shown by :

1

2

3

4

Explanation:

C The magnetic field due to current I
inside $(r \ltR)$ and outside $(r \geq R)$ the wire
For $r < R$ (inside the wire)
$B = \frac{μ_{o} I}{2 π r} \Rightarrow B \propto \frac{1}{r}$
For $r \geq R$ (outside the wire)
$B = \frac{μ_{o} Ir}{2 π R^{2}} \Rightarrow B \propto r$

Karnataka CET-2020

Moving Charges & Magnetism

153164
As shown in the figure, a wire is bent to form a Dshaped closed loop, carrying current I, where the curved part is a semi-circle of radius $R$ . The loop is placed in a uniform magnetic field $\vec{B}$ , which is directed into the plane of the paper. The magnetic force felt by the closed loop is

1 zero

2 IRB

3 2 IRB

4

\frac{1}{2} IRB

WB JEE 2020

Moving Charges & Magnetism

153167 When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to current ' $I$ ' flowing through it is $B_{1}$ . If the same wire is turned into four loops to make a circular coil, the magnetic induction at the centre of this coil is ' $B_{2}$ ' for same current then relation between $B_{2}$ and $B_{1}$ is

1

B_{2} = 8 B_{1}

2

B_{2} = 16 B_{1}

3

B_{2} = 4 B_{1}

4

B_{2} = 64 B_{1}

Explanation:

B Perimeter of circular loop,
$2 π r_{1} = l$
$r_{1} = \frac{l}{2 π}$
When, there is four loops
Then Perimeter of all circular loops
$4 \times (2 π r_{2}) = l$
$8 π r_{2} = l$
$r_{2} = \frac{l}{8 π}$
Case (i)- Number of loop $(N_{1}) = 1$
Magnetic induction at the center of circle due to current I.
$B_{1} = \frac{μ_{0} N_{1} I}{2 r_{1}}$
Case (ii)- Number of loop $(N_{2}) = 4$
When same wire is turned into four loops
$B_{2} = \frac{μ_{0} N_{2} I}{2 r_{2}}$
Then, Ratio of the magnetic flux
$\frac{B_{1}}{B_{2}} = \frac{N_{1}}{N_{2}} \times \frac{r_{2}}{r_{1}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{4} \times \frac{\frac{l}{8 π}}{\frac{l}{2 π}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{16}$
$B_{2} = 16 B_{1}$

MHT-CET 2020

Moving Charges & Magnetism

153168 An electron moves in a circular orbit of radius ' $r$ ' with uniform speed ' $v$ '. It produces magnetic field ' $B$ ' at the centre of circle. The magnetic field ' $B$ ' is proportional to

1

\frac{v}{r^{2}}

2

\frac{1}{{vr}^{2}}

3

{vr}^{2}

4

\frac{r^{2}}{v}

Explanation:

A An electron moves in circular orbit with uniform speed $v$ and having radius $r$ .
The time period of electron moving in a circular orbit,
$T = \frac{Circumference of circular path}{Speed}$

$t = \frac{2 π r}{v}$
And current due to electron, $(q = e)$
$i = \frac{q}{t} = \frac{e}{t} (∵ e = it)$
$i = \frac{e}{(\frac{2 π r}{v})}$
$i = \frac{ev}{2 π r}$
The magnetic field at the centre of circle,
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{μ_{0} \times ev}{2 r \times 2 π r} = \frac{μ_{0} ev}{4 π r^{2}}$
$B \propto \frac{v}{r^{2}}$

MHT-CET 2020

Moving Charges & Magnetism

153170 An infinitely long straight conductor is bent into shape as shown in figure. It carries a current $I A$ and the radius of circular loop is $r$ $m$ . The magnetic induction at the centre of the circular loop is

1

\frac{μ_{0} I (π - 1)}{2 π r}

2

\frac{μ_{0} I (π + 1)}{2 π r}

3

\frac{μ_{0} I (2 π - 1)}{2 π r}

4

\frac{μ_{0} I (2 π + 1)}{2 π r}

Explanation:

A Given, radius of circle $= r$ , current $= I A$

Magnetic field due to straight wire -
$B_{Straight} = \frac{μ_{0}}{4 π} \frac{2 I}{r}$
(direction of field will be outward by right hand thumb rule)
Magnetic field due to circular wire-
$B_{Circle} = \frac{μ_{0}}{4 π} \frac{2 π I}{r}$
(direction of field will be inward)
Since, both the field are in opposite direction
$∴$ Net magnetic field $(B_{net}) = B_{circle} - B_{straight}$
$B_{net} = \frac{μ_{0}}{4 π} \frac{2 π I}{r} - \frac{μ_{0}}{4 π} \frac{2 I}{r}$
$B_{net} = \frac{μ_{0}}{4 π} \cdot \frac{2 I}{r} (π - 1) = \frac{μ_{0} I}{2 π r} (π - 1)$

AP EAMCET (22.09.2020) Shift-II

Moving Charges & Magnetism

153166 A long cylindrical wire of radius $R$ carries a uniform current $I$ flowing through it. The variation of magnetic field with distance $r$ from the axis of the wire is shown by :

1

2

3

4

Explanation:

C The magnetic field due to current I
inside $(r \ltR)$ and outside $(r \geq R)$ the wire
For $r < R$ (inside the wire)
$B = \frac{μ_{o} I}{2 π r} \Rightarrow B \propto \frac{1}{r}$
For $r \geq R$ (outside the wire)
$B = \frac{μ_{o} Ir}{2 π R^{2}} \Rightarrow B \propto r$

Karnataka CET-2020

Moving Charges & Magnetism

153164
As shown in the figure, a wire is bent to form a Dshaped closed loop, carrying current I, where the curved part is a semi-circle of radius $R$ . The loop is placed in a uniform magnetic field $\vec{B}$ , which is directed into the plane of the paper. The magnetic force felt by the closed loop is

1 zero

2 IRB

3 2 IRB

4

\frac{1}{2} IRB

WB JEE 2020

Moving Charges & Magnetism

153167 When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to current ' $I$ ' flowing through it is $B_{1}$ . If the same wire is turned into four loops to make a circular coil, the magnetic induction at the centre of this coil is ' $B_{2}$ ' for same current then relation between $B_{2}$ and $B_{1}$ is

1

B_{2} = 8 B_{1}

2

B_{2} = 16 B_{1}

3

B_{2} = 4 B_{1}

4

B_{2} = 64 B_{1}

Explanation:

B Perimeter of circular loop,
$2 π r_{1} = l$
$r_{1} = \frac{l}{2 π}$
When, there is four loops
Then Perimeter of all circular loops
$4 \times (2 π r_{2}) = l$
$8 π r_{2} = l$
$r_{2} = \frac{l}{8 π}$
Case (i)- Number of loop $(N_{1}) = 1$
Magnetic induction at the center of circle due to current I.
$B_{1} = \frac{μ_{0} N_{1} I}{2 r_{1}}$
Case (ii)- Number of loop $(N_{2}) = 4$
When same wire is turned into four loops
$B_{2} = \frac{μ_{0} N_{2} I}{2 r_{2}}$
Then, Ratio of the magnetic flux
$\frac{B_{1}}{B_{2}} = \frac{N_{1}}{N_{2}} \times \frac{r_{2}}{r_{1}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{4} \times \frac{\frac{l}{8 π}}{\frac{l}{2 π}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{16}$
$B_{2} = 16 B_{1}$

MHT-CET 2020

Moving Charges & Magnetism

153168 An electron moves in a circular orbit of radius ' $r$ ' with uniform speed ' $v$ '. It produces magnetic field ' $B$ ' at the centre of circle. The magnetic field ' $B$ ' is proportional to

1

\frac{v}{r^{2}}

2

\frac{1}{{vr}^{2}}

3

{vr}^{2}

4

\frac{r^{2}}{v}

Explanation:

A An electron moves in circular orbit with uniform speed $v$ and having radius $r$ .
The time period of electron moving in a circular orbit,
$T = \frac{Circumference of circular path}{Speed}$

$t = \frac{2 π r}{v}$
And current due to electron, $(q = e)$
$i = \frac{q}{t} = \frac{e}{t} (∵ e = it)$
$i = \frac{e}{(\frac{2 π r}{v})}$
$i = \frac{ev}{2 π r}$
The magnetic field at the centre of circle,
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{μ_{0} \times ev}{2 r \times 2 π r} = \frac{μ_{0} ev}{4 π r^{2}}$
$B \propto \frac{v}{r^{2}}$

MHT-CET 2020

Moving Charges & Magnetism

153170 An infinitely long straight conductor is bent into shape as shown in figure. It carries a current $I A$ and the radius of circular loop is $r$ $m$ . The magnetic induction at the centre of the circular loop is

1

\frac{μ_{0} I (π - 1)}{2 π r}

2

\frac{μ_{0} I (π + 1)}{2 π r}

3

\frac{μ_{0} I (2 π - 1)}{2 π r}

4

\frac{μ_{0} I (2 π + 1)}{2 π r}

Explanation:

A Given, radius of circle $= r$ , current $= I A$

Magnetic field due to straight wire -
$B_{Straight} = \frac{μ_{0}}{4 π} \frac{2 I}{r}$
(direction of field will be outward by right hand thumb rule)
Magnetic field due to circular wire-
$B_{Circle} = \frac{μ_{0}}{4 π} \frac{2 π I}{r}$
(direction of field will be inward)
Since, both the field are in opposite direction
$∴$ Net magnetic field $(B_{net}) = B_{circle} - B_{straight}$
$B_{net} = \frac{μ_{0}}{4 π} \frac{2 π I}{r} - \frac{μ_{0}}{4 π} \frac{2 I}{r}$
$B_{net} = \frac{μ_{0}}{4 π} \cdot \frac{2 I}{r} (π - 1) = \frac{μ_{0} I}{2 π r} (π - 1)$

AP EAMCET (22.09.2020) Shift-II

Moving Charges & Magnetism

153166 A long cylindrical wire of radius $R$ carries a uniform current $I$ flowing through it. The variation of magnetic field with distance $r$ from the axis of the wire is shown by :

1

2

3

4

Explanation:

C The magnetic field due to current I
inside $(r \ltR)$ and outside $(r \geq R)$ the wire
For $r < R$ (inside the wire)
$B = \frac{μ_{o} I}{2 π r} \Rightarrow B \propto \frac{1}{r}$
For $r \geq R$ (outside the wire)
$B = \frac{μ_{o} Ir}{2 π R^{2}} \Rightarrow B \propto r$

Karnataka CET-2020

Moving Charges & Magnetism

153164
As shown in the figure, a wire is bent to form a Dshaped closed loop, carrying current I, where the curved part is a semi-circle of radius $R$ . The loop is placed in a uniform magnetic field $\vec{B}$ , which is directed into the plane of the paper. The magnetic force felt by the closed loop is

1 zero

2 IRB

3 2 IRB

4

\frac{1}{2} IRB

WB JEE 2020

Moving Charges & Magnetism

153167 When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to current ' $I$ ' flowing through it is $B_{1}$ . If the same wire is turned into four loops to make a circular coil, the magnetic induction at the centre of this coil is ' $B_{2}$ ' for same current then relation between $B_{2}$ and $B_{1}$ is

1

B_{2} = 8 B_{1}

2

B_{2} = 16 B_{1}

3

B_{2} = 4 B_{1}

4

B_{2} = 64 B_{1}

Explanation:

B Perimeter of circular loop,
$2 π r_{1} = l$
$r_{1} = \frac{l}{2 π}$
When, there is four loops
Then Perimeter of all circular loops
$4 \times (2 π r_{2}) = l$
$8 π r_{2} = l$
$r_{2} = \frac{l}{8 π}$
Case (i)- Number of loop $(N_{1}) = 1$
Magnetic induction at the center of circle due to current I.
$B_{1} = \frac{μ_{0} N_{1} I}{2 r_{1}}$
Case (ii)- Number of loop $(N_{2}) = 4$
When same wire is turned into four loops
$B_{2} = \frac{μ_{0} N_{2} I}{2 r_{2}}$
Then, Ratio of the magnetic flux
$\frac{B_{1}}{B_{2}} = \frac{N_{1}}{N_{2}} \times \frac{r_{2}}{r_{1}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{4} \times \frac{\frac{l}{8 π}}{\frac{l}{2 π}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{16}$
$B_{2} = 16 B_{1}$

MHT-CET 2020

Moving Charges & Magnetism

153168 An electron moves in a circular orbit of radius ' $r$ ' with uniform speed ' $v$ '. It produces magnetic field ' $B$ ' at the centre of circle. The magnetic field ' $B$ ' is proportional to

1

\frac{v}{r^{2}}

2

\frac{1}{{vr}^{2}}

3

{vr}^{2}

4

\frac{r^{2}}{v}

Explanation:

A An electron moves in circular orbit with uniform speed $v$ and having radius $r$ .
The time period of electron moving in a circular orbit,
$T = \frac{Circumference of circular path}{Speed}$

$t = \frac{2 π r}{v}$
And current due to electron, $(q = e)$
$i = \frac{q}{t} = \frac{e}{t} (∵ e = it)$
$i = \frac{e}{(\frac{2 π r}{v})}$
$i = \frac{ev}{2 π r}$
The magnetic field at the centre of circle,
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{μ_{0} \times ev}{2 r \times 2 π r} = \frac{μ_{0} ev}{4 π r^{2}}$
$B \propto \frac{v}{r^{2}}$

MHT-CET 2020

Moving Charges & Magnetism

153170 An infinitely long straight conductor is bent into shape as shown in figure. It carries a current $I A$ and the radius of circular loop is $r$ $m$ . The magnetic induction at the centre of the circular loop is

1

\frac{μ_{0} I (π - 1)}{2 π r}

2

\frac{μ_{0} I (π + 1)}{2 π r}

3

\frac{μ_{0} I (2 π - 1)}{2 π r}

4

\frac{μ_{0} I (2 π + 1)}{2 π r}

Explanation:

A Given, radius of circle $= r$ , current $= I A$

Magnetic field due to straight wire -
$B_{Straight} = \frac{μ_{0}}{4 π} \frac{2 I}{r}$
(direction of field will be outward by right hand thumb rule)
Magnetic field due to circular wire-
$B_{Circle} = \frac{μ_{0}}{4 π} \frac{2 π I}{r}$
(direction of field will be inward)
Since, both the field are in opposite direction
$∴$ Net magnetic field $(B_{net}) = B_{circle} - B_{straight}$
$B_{net} = \frac{μ_{0}}{4 π} \frac{2 π I}{r} - \frac{μ_{0}}{4 π} \frac{2 I}{r}$
$B_{net} = \frac{μ_{0}}{4 π} \cdot \frac{2 I}{r} (π - 1) = \frac{μ_{0} I}{2 π r} (π - 1)$

AP EAMCET (22.09.2020) Shift-II

Moving Charges & Magnetism

153166 A long cylindrical wire of radius $R$ carries a uniform current $I$ flowing through it. The variation of magnetic field with distance $r$ from the axis of the wire is shown by :

1

2

3

4

Explanation:

C The magnetic field due to current I
inside $(r \ltR)$ and outside $(r \geq R)$ the wire
For $r < R$ (inside the wire)
$B = \frac{μ_{o} I}{2 π r} \Rightarrow B \propto \frac{1}{r}$
For $r \geq R$ (outside the wire)
$B = \frac{μ_{o} Ir}{2 π R^{2}} \Rightarrow B \propto r$

Karnataka CET-2020

Moving Charges & Magnetism

153164
As shown in the figure, a wire is bent to form a Dshaped closed loop, carrying current I, where the curved part is a semi-circle of radius $R$ . The loop is placed in a uniform magnetic field $\vec{B}$ , which is directed into the plane of the paper. The magnetic force felt by the closed loop is

1 zero

2 IRB

3 2 IRB

4

\frac{1}{2} IRB

WB JEE 2020

Moving Charges & Magnetism

153167 When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to current ' $I$ ' flowing through it is $B_{1}$ . If the same wire is turned into four loops to make a circular coil, the magnetic induction at the centre of this coil is ' $B_{2}$ ' for same current then relation between $B_{2}$ and $B_{1}$ is

1

B_{2} = 8 B_{1}

2

B_{2} = 16 B_{1}

3

B_{2} = 4 B_{1}

4

B_{2} = 64 B_{1}

Explanation:

B Perimeter of circular loop,
$2 π r_{1} = l$
$r_{1} = \frac{l}{2 π}$
When, there is four loops
Then Perimeter of all circular loops
$4 \times (2 π r_{2}) = l$
$8 π r_{2} = l$
$r_{2} = \frac{l}{8 π}$
Case (i)- Number of loop $(N_{1}) = 1$
Magnetic induction at the center of circle due to current I.
$B_{1} = \frac{μ_{0} N_{1} I}{2 r_{1}}$
Case (ii)- Number of loop $(N_{2}) = 4$
When same wire is turned into four loops
$B_{2} = \frac{μ_{0} N_{2} I}{2 r_{2}}$
Then, Ratio of the magnetic flux
$\frac{B_{1}}{B_{2}} = \frac{N_{1}}{N_{2}} \times \frac{r_{2}}{r_{1}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{4} \times \frac{\frac{l}{8 π}}{\frac{l}{2 π}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{16}$
$B_{2} = 16 B_{1}$

MHT-CET 2020

Moving Charges & Magnetism

153168 An electron moves in a circular orbit of radius ' $r$ ' with uniform speed ' $v$ '. It produces magnetic field ' $B$ ' at the centre of circle. The magnetic field ' $B$ ' is proportional to

1

\frac{v}{r^{2}}

2

\frac{1}{{vr}^{2}}

3

{vr}^{2}

4

\frac{r^{2}}{v}

Explanation:

A An electron moves in circular orbit with uniform speed $v$ and having radius $r$ .
The time period of electron moving in a circular orbit,
$T = \frac{Circumference of circular path}{Speed}$

$t = \frac{2 π r}{v}$
And current due to electron, $(q = e)$
$i = \frac{q}{t} = \frac{e}{t} (∵ e = it)$
$i = \frac{e}{(\frac{2 π r}{v})}$
$i = \frac{ev}{2 π r}$
The magnetic field at the centre of circle,
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{μ_{0} \times ev}{2 r \times 2 π r} = \frac{μ_{0} ev}{4 π r^{2}}$
$B \propto \frac{v}{r^{2}}$

MHT-CET 2020

Moving Charges & Magnetism

153170 An infinitely long straight conductor is bent into shape as shown in figure. It carries a current $I A$ and the radius of circular loop is $r$ $m$ . The magnetic induction at the centre of the circular loop is

1

\frac{μ_{0} I (π - 1)}{2 π r}

2

\frac{μ_{0} I (π + 1)}{2 π r}

3

\frac{μ_{0} I (2 π - 1)}{2 π r}

4

\frac{μ_{0} I (2 π + 1)}{2 π r}

Explanation:

A Given, radius of circle $= r$ , current $= I A$

Magnetic field due to straight wire -
$B_{Straight} = \frac{μ_{0}}{4 π} \frac{2 I}{r}$
(direction of field will be outward by right hand thumb rule)
Magnetic field due to circular wire-
$B_{Circle} = \frac{μ_{0}}{4 π} \frac{2 π I}{r}$
(direction of field will be inward)
Since, both the field are in opposite direction
$∴$ Net magnetic field $(B_{net}) = B_{circle} - B_{straight}$
$B_{net} = \frac{μ_{0}}{4 π} \frac{2 π I}{r} - \frac{μ_{0}}{4 π} \frac{2 I}{r}$
$B_{net} = \frac{μ_{0}}{4 π} \cdot \frac{2 I}{r} (π - 1) = \frac{μ_{0} I}{2 π r} (π - 1)$

AP EAMCET (22.09.2020) Shift-II

Moving Charges & Magnetism

153166 A long cylindrical wire of radius $R$ carries a uniform current $I$ flowing through it. The variation of magnetic field with distance $r$ from the axis of the wire is shown by :

1

2

3

4

Explanation:

C The magnetic field due to current I
inside $(r \ltR)$ and outside $(r \geq R)$ the wire
For $r < R$ (inside the wire)
$B = \frac{μ_{o} I}{2 π r} \Rightarrow B \propto \frac{1}{r}$
For $r \geq R$ (outside the wire)
$B = \frac{μ_{o} Ir}{2 π R^{2}} \Rightarrow B \propto r$

Karnataka CET-2020

Moving Charges & Magnetism

153164
As shown in the figure, a wire is bent to form a Dshaped closed loop, carrying current I, where the curved part is a semi-circle of radius $R$ . The loop is placed in a uniform magnetic field $\vec{B}$ , which is directed into the plane of the paper. The magnetic force felt by the closed loop is

1 zero

2 IRB

3 2 IRB

4

\frac{1}{2} IRB

WB JEE 2020

Moving Charges & Magnetism

153167 When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to current ' $I$ ' flowing through it is $B_{1}$ . If the same wire is turned into four loops to make a circular coil, the magnetic induction at the centre of this coil is ' $B_{2}$ ' for same current then relation between $B_{2}$ and $B_{1}$ is

1

B_{2} = 8 B_{1}

2

B_{2} = 16 B_{1}

3

B_{2} = 4 B_{1}

4

B_{2} = 64 B_{1}

Explanation:

B Perimeter of circular loop,
$2 π r_{1} = l$
$r_{1} = \frac{l}{2 π}$
When, there is four loops
Then Perimeter of all circular loops
$4 \times (2 π r_{2}) = l$
$8 π r_{2} = l$
$r_{2} = \frac{l}{8 π}$
Case (i)- Number of loop $(N_{1}) = 1$
Magnetic induction at the center of circle due to current I.
$B_{1} = \frac{μ_{0} N_{1} I}{2 r_{1}}$
Case (ii)- Number of loop $(N_{2}) = 4$
When same wire is turned into four loops
$B_{2} = \frac{μ_{0} N_{2} I}{2 r_{2}}$
Then, Ratio of the magnetic flux
$\frac{B_{1}}{B_{2}} = \frac{N_{1}}{N_{2}} \times \frac{r_{2}}{r_{1}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{4} \times \frac{\frac{l}{8 π}}{\frac{l}{2 π}}$
$\frac{B_{1}}{B_{2}} = \frac{1}{16}$
$B_{2} = 16 B_{1}$

MHT-CET 2020

Moving Charges & Magnetism

153168 An electron moves in a circular orbit of radius ' $r$ ' with uniform speed ' $v$ '. It produces magnetic field ' $B$ ' at the centre of circle. The magnetic field ' $B$ ' is proportional to

1

\frac{v}{r^{2}}

2

\frac{1}{{vr}^{2}}

3

{vr}^{2}

4

\frac{r^{2}}{v}

Explanation:

A An electron moves in circular orbit with uniform speed $v$ and having radius $r$ .
The time period of electron moving in a circular orbit,
$T = \frac{Circumference of circular path}{Speed}$

$t = \frac{2 π r}{v}$
And current due to electron, $(q = e)$
$i = \frac{q}{t} = \frac{e}{t} (∵ e = it)$
$i = \frac{e}{(\frac{2 π r}{v})}$
$i = \frac{ev}{2 π r}$
The magnetic field at the centre of circle,
$B = \frac{μ_{0} i}{2 r}$
$B = \frac{μ_{0} \times ev}{2 r \times 2 π r} = \frac{μ_{0} ev}{4 π r^{2}}$
$B \propto \frac{v}{r^{2}}$

MHT-CET 2020

Moving Charges & Magnetism

153170 An infinitely long straight conductor is bent into shape as shown in figure. It carries a current $I A$ and the radius of circular loop is $r$ $m$ . The magnetic induction at the centre of the circular loop is

1

\frac{μ_{0} I (π - 1)}{2 π r}

2

\frac{μ_{0} I (π + 1)}{2 π r}

3

\frac{μ_{0} I (2 π - 1)}{2 π r}

4

\frac{μ_{0} I (2 π + 1)}{2 π r}

Explanation:

A Given, radius of circle $= r$ , current $= I A$

Magnetic field due to straight wire -
$B_{Straight} = \frac{μ_{0}}{4 π} \frac{2 I}{r}$
(direction of field will be outward by right hand thumb rule)
Magnetic field due to circular wire-
$B_{Circle} = \frac{μ_{0}}{4 π} \frac{2 π I}{r}$
(direction of field will be inward)
Since, both the field are in opposite direction
$∴$ Net magnetic field $(B_{net}) = B_{circle} - B_{straight}$
$B_{net} = \frac{μ_{0}}{4 π} \frac{2 π I}{r} - \frac{μ_{0}}{4 π} \frac{2 I}{r}$
$B_{net} = \frac{μ_{0}}{4 π} \cdot \frac{2 I}{r} (π - 1) = \frac{μ_{0} I}{2 π r} (π - 1)$

AP EAMCET (22.09.2020) Shift-II