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Moving Charges & Magnetism

153218 The magnitude of a magnetic field at the centre of a circular coil of radius $R$, having $\mathbf{N}$ turns and carrying a current $I$ can be doubled by changing

1 I to $2 \mathrm{I}$ and $\mathrm{N}$ to $2 \mathrm{~N}$ keeping $\mathrm{R}$ unchanged

2 $\mathrm{N}$ to $\mathrm{N} / 2$ and keeping I and $\mathrm{R}$ unchanged

3 $\mathrm{N}$ to $2 \mathrm{~N}$ and $\mathrm{R}$ to $2 \mathrm{R}$ keeping I unchanged

4 $\mathrm{R}$ to $2 \mathrm{R}$ and I to $2 \mathrm{I}$ keeping $\mathrm{N}$ unchanged

5 1 to $2 \mathrm{I}$ and keeping $\mathrm{N}$ and $\mathrm{R}$ unchanged

Kerala CEE -2018

Moving Charges & Magnetism

153219 An electron is moving with a velocity $2 \times 10^{6}$ $\mathrm{m} / \mathrm{s}$ along positive $\mathrm{x}$-direction in the uniform electric field of $8 \times 10^{7} \mathrm{~V} / \mathrm{m}$ applied along positive y-direction. The magnitude and direction of a uniform magnetic field (in tesla) that will cause the electrons to move undeviated along its original path is

1 40 in - ve z-direction

2 40 in + ve z-direction

3 4 in + ve $z$-direction

4 4 in - ve z-direction

5 8 in + ve z-direction

Kerala CEE -2018

Moving Charges & Magnetism

153220 A rectangular conducting loop of length $4 \sqrt{2} \mathrm{~m}$ and breadth $4 \mathrm{~m}$ carrying a current of $5 \mathrm{~A}$ in the anti-clockwise direction is placed in the $x y$ plane. The magnitude of the magnetic induction field vector $B$ at the intersection of the diagonal is $\quad\left(\right.$ use $\mu_{0}=4 \times 10^{-7} \mathrm{NA}^{-2}$ )

1 $1.2 \times 10^{-6} \mathrm{~T}$

2 $1.2 \times 10^{-5} \mathrm{~T}$

3 $2.4 \times 10^{-6} \mathrm{~T}$

4 $2.4 \times 10^{-5} \mathrm{~T}$

5 $1.2 \times 10^{-7} \mathrm{~T}$

Explanation:

A Given, current (I) $=5 \mathrm{~A}$
According to question,

Since, $\quad \mathrm{AB}=4 \mathrm{~m} \& \mathrm{AD}=4 \sqrt{2}$
For rectangular both diagonals are same and bisect each other
$\because \quad \mathrm{AC}=\mathrm{BD}=\sqrt{(4 \sqrt{2})^{2}+4^{2}}=\sqrt{48}$
$\mathrm{AC}=\mathrm{BD}=4 \sqrt{3}$
$\because \quad \mathrm{AP}=\mathrm{BP}=\mathrm{CP}=\mathrm{DP}=\frac{\mathrm{AC}}{2}=\frac{4 \sqrt{3}}{2}=2 \sqrt{3}$
$\sin \alpha_{1}=\frac{\mathrm{DF}}{\mathrm{DP}}=\frac{2 \sqrt{2}}{2 \sqrt{3}}$
$\sin \alpha_{1}=0.816$
Similarly, $\sin \alpha_{2}=0.816$
$\sin \alpha_{3}=\frac{\mathrm{AE}}{\mathrm{AP}}=\frac{2}{2 \sqrt{3}}$
And $\quad \sin \alpha_{3}=0.57$
And $\quad \sin \alpha_{4}=0.57$
Magnetic field due to conductor $\mathrm{AD}$
$\mathrm{r}_{1}=2 \mathrm{~m}$
$\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}_{1}}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
Putting this value in equation (i), we get -
$\mathrm{B}_{1} =\frac{\mu_{0}}{4 \pi} \times \frac{5}{2}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
$\mathrm{B}_{1} =10^{-7} \times \frac{5}{2}(0.816+0.816)$
$\mathrm{B}_{1} =4.08 \times 10^{-7} \mathrm{~T}$
Similarly, magnetic field due to conductor $\mathrm{BC}$
$\mathrm{B}_{2}=4.08 \times 10^{-7} \mathrm{~T}$
Magnetic field due to conductor $\mathrm{AB}$
$\mathrm{r}_{2}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}$
$\mathrm{~B}_{3}=\frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}_{2}}\left[\sin \alpha_{3}+\sin \alpha_{4}\right]$
$\mathrm{B}_{3}=10^{-7} \times \frac{5}{2 \sqrt{2}}[0.57+0.57]$
$\mathrm{B}_3=2.04 \times 10^{-7} \mathrm{~T}$
Similarly field along conductor $\mathrm{CD}$
$\mathrm{B}_4=2.04 \times 10^{-7} \mathrm{~T}$
Resultant magnetic field
$\mathrm{B} =\mathrm{B}_1+\mathrm{B}_2+\mathrm{B}_3+\mathrm{B}_4$
$=[4.08+4.08+2.04+2.04] \times 10^{-7}$
$\therefore \quad \mathrm{B} =12.29 \times 10^{-7} \mathrm{~T}$
$\therefore \quad \mathrm{B} =1.2 \times 10^{-6} \mathrm{~T}$

Kerala CEE -2018

Moving Charges & Magnetism

153221 A long straight wire carrying electric current $i$ is bent at its mid-point to form an angle of $45^{\circ}$ as shown in the figure. Magnetic field at a point $P$ at a distance $d$ from the point $Q$ of bending is

1 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}[\sqrt{2}-1]$

2 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}[\sqrt{2}-1]$

3 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}$

4 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}$

AP EAMCET (23.04.2018) Shift-1

Moving Charges & Magnetism

153218 The magnitude of a magnetic field at the centre of a circular coil of radius $R$, having $\mathbf{N}$ turns and carrying a current $I$ can be doubled by changing

1 I to $2 \mathrm{I}$ and $\mathrm{N}$ to $2 \mathrm{~N}$ keeping $\mathrm{R}$ unchanged

2 $\mathrm{N}$ to $\mathrm{N} / 2$ and keeping I and $\mathrm{R}$ unchanged

3 $\mathrm{N}$ to $2 \mathrm{~N}$ and $\mathrm{R}$ to $2 \mathrm{R}$ keeping I unchanged

4 $\mathrm{R}$ to $2 \mathrm{R}$ and I to $2 \mathrm{I}$ keeping $\mathrm{N}$ unchanged

5 1 to $2 \mathrm{I}$ and keeping $\mathrm{N}$ and $\mathrm{R}$ unchanged

Kerala CEE -2018

Moving Charges & Magnetism

153219 An electron is moving with a velocity $2 \times 10^{6}$ $\mathrm{m} / \mathrm{s}$ along positive $\mathrm{x}$-direction in the uniform electric field of $8 \times 10^{7} \mathrm{~V} / \mathrm{m}$ applied along positive y-direction. The magnitude and direction of a uniform magnetic field (in tesla) that will cause the electrons to move undeviated along its original path is

1 40 in - ve z-direction

2 40 in + ve z-direction

3 4 in + ve $z$-direction

4 4 in - ve z-direction

5 8 in + ve z-direction

Kerala CEE -2018

Moving Charges & Magnetism

153220 A rectangular conducting loop of length $4 \sqrt{2} \mathrm{~m}$ and breadth $4 \mathrm{~m}$ carrying a current of $5 \mathrm{~A}$ in the anti-clockwise direction is placed in the $x y$ plane. The magnitude of the magnetic induction field vector $B$ at the intersection of the diagonal is $\quad\left(\right.$ use $\mu_{0}=4 \times 10^{-7} \mathrm{NA}^{-2}$ )

1 $1.2 \times 10^{-6} \mathrm{~T}$

2 $1.2 \times 10^{-5} \mathrm{~T}$

3 $2.4 \times 10^{-6} \mathrm{~T}$

4 $2.4 \times 10^{-5} \mathrm{~T}$

5 $1.2 \times 10^{-7} \mathrm{~T}$

Explanation:

A Given, current (I) $=5 \mathrm{~A}$
According to question,

Since, $\quad \mathrm{AB}=4 \mathrm{~m} \& \mathrm{AD}=4 \sqrt{2}$
For rectangular both diagonals are same and bisect each other
$\because \quad \mathrm{AC}=\mathrm{BD}=\sqrt{(4 \sqrt{2})^{2}+4^{2}}=\sqrt{48}$
$\mathrm{AC}=\mathrm{BD}=4 \sqrt{3}$
$\because \quad \mathrm{AP}=\mathrm{BP}=\mathrm{CP}=\mathrm{DP}=\frac{\mathrm{AC}}{2}=\frac{4 \sqrt{3}}{2}=2 \sqrt{3}$
$\sin \alpha_{1}=\frac{\mathrm{DF}}{\mathrm{DP}}=\frac{2 \sqrt{2}}{2 \sqrt{3}}$
$\sin \alpha_{1}=0.816$
Similarly, $\sin \alpha_{2}=0.816$
$\sin \alpha_{3}=\frac{\mathrm{AE}}{\mathrm{AP}}=\frac{2}{2 \sqrt{3}}$
And $\quad \sin \alpha_{3}=0.57$
And $\quad \sin \alpha_{4}=0.57$
Magnetic field due to conductor $\mathrm{AD}$
$\mathrm{r}_{1}=2 \mathrm{~m}$
$\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}_{1}}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
Putting this value in equation (i), we get -
$\mathrm{B}_{1} =\frac{\mu_{0}}{4 \pi} \times \frac{5}{2}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
$\mathrm{B}_{1} =10^{-7} \times \frac{5}{2}(0.816+0.816)$
$\mathrm{B}_{1} =4.08 \times 10^{-7} \mathrm{~T}$
Similarly, magnetic field due to conductor $\mathrm{BC}$
$\mathrm{B}_{2}=4.08 \times 10^{-7} \mathrm{~T}$
Magnetic field due to conductor $\mathrm{AB}$
$\mathrm{r}_{2}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}$
$\mathrm{~B}_{3}=\frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}_{2}}\left[\sin \alpha_{3}+\sin \alpha_{4}\right]$
$\mathrm{B}_{3}=10^{-7} \times \frac{5}{2 \sqrt{2}}[0.57+0.57]$
$\mathrm{B}_3=2.04 \times 10^{-7} \mathrm{~T}$
Similarly field along conductor $\mathrm{CD}$
$\mathrm{B}_4=2.04 \times 10^{-7} \mathrm{~T}$
Resultant magnetic field
$\mathrm{B} =\mathrm{B}_1+\mathrm{B}_2+\mathrm{B}_3+\mathrm{B}_4$
$=[4.08+4.08+2.04+2.04] \times 10^{-7}$
$\therefore \quad \mathrm{B} =12.29 \times 10^{-7} \mathrm{~T}$
$\therefore \quad \mathrm{B} =1.2 \times 10^{-6} \mathrm{~T}$

Kerala CEE -2018

Moving Charges & Magnetism

153221 A long straight wire carrying electric current $i$ is bent at its mid-point to form an angle of $45^{\circ}$ as shown in the figure. Magnetic field at a point $P$ at a distance $d$ from the point $Q$ of bending is

1 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}[\sqrt{2}-1]$

2 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}[\sqrt{2}-1]$

3 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}$

4 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}$

AP EAMCET (23.04.2018) Shift-1

Moving Charges & Magnetism

153218 The magnitude of a magnetic field at the centre of a circular coil of radius $R$, having $\mathbf{N}$ turns and carrying a current $I$ can be doubled by changing

1 I to $2 \mathrm{I}$ and $\mathrm{N}$ to $2 \mathrm{~N}$ keeping $\mathrm{R}$ unchanged

2 $\mathrm{N}$ to $\mathrm{N} / 2$ and keeping I and $\mathrm{R}$ unchanged

3 $\mathrm{N}$ to $2 \mathrm{~N}$ and $\mathrm{R}$ to $2 \mathrm{R}$ keeping I unchanged

4 $\mathrm{R}$ to $2 \mathrm{R}$ and I to $2 \mathrm{I}$ keeping $\mathrm{N}$ unchanged

5 1 to $2 \mathrm{I}$ and keeping $\mathrm{N}$ and $\mathrm{R}$ unchanged

Kerala CEE -2018

Moving Charges & Magnetism

153219 An electron is moving with a velocity $2 \times 10^{6}$ $\mathrm{m} / \mathrm{s}$ along positive $\mathrm{x}$-direction in the uniform electric field of $8 \times 10^{7} \mathrm{~V} / \mathrm{m}$ applied along positive y-direction. The magnitude and direction of a uniform magnetic field (in tesla) that will cause the electrons to move undeviated along its original path is

1 40 in - ve z-direction

2 40 in + ve z-direction

3 4 in + ve $z$-direction

4 4 in - ve z-direction

5 8 in + ve z-direction

Kerala CEE -2018

Moving Charges & Magnetism

153220 A rectangular conducting loop of length $4 \sqrt{2} \mathrm{~m}$ and breadth $4 \mathrm{~m}$ carrying a current of $5 \mathrm{~A}$ in the anti-clockwise direction is placed in the $x y$ plane. The magnitude of the magnetic induction field vector $B$ at the intersection of the diagonal is $\quad\left(\right.$ use $\mu_{0}=4 \times 10^{-7} \mathrm{NA}^{-2}$ )

1 $1.2 \times 10^{-6} \mathrm{~T}$

2 $1.2 \times 10^{-5} \mathrm{~T}$

3 $2.4 \times 10^{-6} \mathrm{~T}$

4 $2.4 \times 10^{-5} \mathrm{~T}$

5 $1.2 \times 10^{-7} \mathrm{~T}$

Explanation:

A Given, current (I) $=5 \mathrm{~A}$
According to question,

Since, $\quad \mathrm{AB}=4 \mathrm{~m} \& \mathrm{AD}=4 \sqrt{2}$
For rectangular both diagonals are same and bisect each other
$\because \quad \mathrm{AC}=\mathrm{BD}=\sqrt{(4 \sqrt{2})^{2}+4^{2}}=\sqrt{48}$
$\mathrm{AC}=\mathrm{BD}=4 \sqrt{3}$
$\because \quad \mathrm{AP}=\mathrm{BP}=\mathrm{CP}=\mathrm{DP}=\frac{\mathrm{AC}}{2}=\frac{4 \sqrt{3}}{2}=2 \sqrt{3}$
$\sin \alpha_{1}=\frac{\mathrm{DF}}{\mathrm{DP}}=\frac{2 \sqrt{2}}{2 \sqrt{3}}$
$\sin \alpha_{1}=0.816$
Similarly, $\sin \alpha_{2}=0.816$
$\sin \alpha_{3}=\frac{\mathrm{AE}}{\mathrm{AP}}=\frac{2}{2 \sqrt{3}}$
And $\quad \sin \alpha_{3}=0.57$
And $\quad \sin \alpha_{4}=0.57$
Magnetic field due to conductor $\mathrm{AD}$
$\mathrm{r}_{1}=2 \mathrm{~m}$
$\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}_{1}}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
Putting this value in equation (i), we get -
$\mathrm{B}_{1} =\frac{\mu_{0}}{4 \pi} \times \frac{5}{2}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
$\mathrm{B}_{1} =10^{-7} \times \frac{5}{2}(0.816+0.816)$
$\mathrm{B}_{1} =4.08 \times 10^{-7} \mathrm{~T}$
Similarly, magnetic field due to conductor $\mathrm{BC}$
$\mathrm{B}_{2}=4.08 \times 10^{-7} \mathrm{~T}$
Magnetic field due to conductor $\mathrm{AB}$
$\mathrm{r}_{2}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}$
$\mathrm{~B}_{3}=\frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}_{2}}\left[\sin \alpha_{3}+\sin \alpha_{4}\right]$
$\mathrm{B}_{3}=10^{-7} \times \frac{5}{2 \sqrt{2}}[0.57+0.57]$
$\mathrm{B}_3=2.04 \times 10^{-7} \mathrm{~T}$
Similarly field along conductor $\mathrm{CD}$
$\mathrm{B}_4=2.04 \times 10^{-7} \mathrm{~T}$
Resultant magnetic field
$\mathrm{B} =\mathrm{B}_1+\mathrm{B}_2+\mathrm{B}_3+\mathrm{B}_4$
$=[4.08+4.08+2.04+2.04] \times 10^{-7}$
$\therefore \quad \mathrm{B} =12.29 \times 10^{-7} \mathrm{~T}$
$\therefore \quad \mathrm{B} =1.2 \times 10^{-6} \mathrm{~T}$

Kerala CEE -2018

Moving Charges & Magnetism

153221 A long straight wire carrying electric current $i$ is bent at its mid-point to form an angle of $45^{\circ}$ as shown in the figure. Magnetic field at a point $P$ at a distance $d$ from the point $Q$ of bending is

1 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}[\sqrt{2}-1]$

2 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}[\sqrt{2}-1]$

3 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}$

4 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}$

AP EAMCET (23.04.2018) Shift-1

Moving Charges & Magnetism

153218 The magnitude of a magnetic field at the centre of a circular coil of radius $R$, having $\mathbf{N}$ turns and carrying a current $I$ can be doubled by changing

1 I to $2 \mathrm{I}$ and $\mathrm{N}$ to $2 \mathrm{~N}$ keeping $\mathrm{R}$ unchanged

2 $\mathrm{N}$ to $\mathrm{N} / 2$ and keeping I and $\mathrm{R}$ unchanged

3 $\mathrm{N}$ to $2 \mathrm{~N}$ and $\mathrm{R}$ to $2 \mathrm{R}$ keeping I unchanged

4 $\mathrm{R}$ to $2 \mathrm{R}$ and I to $2 \mathrm{I}$ keeping $\mathrm{N}$ unchanged

5 1 to $2 \mathrm{I}$ and keeping $\mathrm{N}$ and $\mathrm{R}$ unchanged

Kerala CEE -2018

Moving Charges & Magnetism

153219 An electron is moving with a velocity $2 \times 10^{6}$ $\mathrm{m} / \mathrm{s}$ along positive $\mathrm{x}$-direction in the uniform electric field of $8 \times 10^{7} \mathrm{~V} / \mathrm{m}$ applied along positive y-direction. The magnitude and direction of a uniform magnetic field (in tesla) that will cause the electrons to move undeviated along its original path is

1 40 in - ve z-direction

2 40 in + ve z-direction

3 4 in + ve $z$-direction

4 4 in - ve z-direction

5 8 in + ve z-direction

Kerala CEE -2018

Moving Charges & Magnetism

153220 A rectangular conducting loop of length $4 \sqrt{2} \mathrm{~m}$ and breadth $4 \mathrm{~m}$ carrying a current of $5 \mathrm{~A}$ in the anti-clockwise direction is placed in the $x y$ plane. The magnitude of the magnetic induction field vector $B$ at the intersection of the diagonal is $\quad\left(\right.$ use $\mu_{0}=4 \times 10^{-7} \mathrm{NA}^{-2}$ )

1 $1.2 \times 10^{-6} \mathrm{~T}$

2 $1.2 \times 10^{-5} \mathrm{~T}$

3 $2.4 \times 10^{-6} \mathrm{~T}$

4 $2.4 \times 10^{-5} \mathrm{~T}$

5 $1.2 \times 10^{-7} \mathrm{~T}$

Explanation:

A Given, current (I) $=5 \mathrm{~A}$
According to question,

Since, $\quad \mathrm{AB}=4 \mathrm{~m} \& \mathrm{AD}=4 \sqrt{2}$
For rectangular both diagonals are same and bisect each other
$\because \quad \mathrm{AC}=\mathrm{BD}=\sqrt{(4 \sqrt{2})^{2}+4^{2}}=\sqrt{48}$
$\mathrm{AC}=\mathrm{BD}=4 \sqrt{3}$
$\because \quad \mathrm{AP}=\mathrm{BP}=\mathrm{CP}=\mathrm{DP}=\frac{\mathrm{AC}}{2}=\frac{4 \sqrt{3}}{2}=2 \sqrt{3}$
$\sin \alpha_{1}=\frac{\mathrm{DF}}{\mathrm{DP}}=\frac{2 \sqrt{2}}{2 \sqrt{3}}$
$\sin \alpha_{1}=0.816$
Similarly, $\sin \alpha_{2}=0.816$
$\sin \alpha_{3}=\frac{\mathrm{AE}}{\mathrm{AP}}=\frac{2}{2 \sqrt{3}}$
And $\quad \sin \alpha_{3}=0.57$
And $\quad \sin \alpha_{4}=0.57$
Magnetic field due to conductor $\mathrm{AD}$
$\mathrm{r}_{1}=2 \mathrm{~m}$
$\mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}_{1}}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
Putting this value in equation (i), we get -
$\mathrm{B}_{1} =\frac{\mu_{0}}{4 \pi} \times \frac{5}{2}\left(\sin \alpha_{1}+\sin \alpha_{2}\right)$
$\mathrm{B}_{1} =10^{-7} \times \frac{5}{2}(0.816+0.816)$
$\mathrm{B}_{1} =4.08 \times 10^{-7} \mathrm{~T}$
Similarly, magnetic field due to conductor $\mathrm{BC}$
$\mathrm{B}_{2}=4.08 \times 10^{-7} \mathrm{~T}$
Magnetic field due to conductor $\mathrm{AB}$
$\mathrm{r}_{2}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}$
$\mathrm{~B}_{3}=\frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{I}}{\mathrm{r}_{2}}\left[\sin \alpha_{3}+\sin \alpha_{4}\right]$
$\mathrm{B}_{3}=10^{-7} \times \frac{5}{2 \sqrt{2}}[0.57+0.57]$
$\mathrm{B}_3=2.04 \times 10^{-7} \mathrm{~T}$
Similarly field along conductor $\mathrm{CD}$
$\mathrm{B}_4=2.04 \times 10^{-7} \mathrm{~T}$
Resultant magnetic field
$\mathrm{B} =\mathrm{B}_1+\mathrm{B}_2+\mathrm{B}_3+\mathrm{B}_4$
$=[4.08+4.08+2.04+2.04] \times 10^{-7}$
$\therefore \quad \mathrm{B} =12.29 \times 10^{-7} \mathrm{~T}$
$\therefore \quad \mathrm{B} =1.2 \times 10^{-6} \mathrm{~T}$

Kerala CEE -2018

Moving Charges & Magnetism

153221 A long straight wire carrying electric current $i$ is bent at its mid-point to form an angle of $45^{\circ}$ as shown in the figure. Magnetic field at a point $P$ at a distance $d$ from the point $Q$ of bending is

1 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}[\sqrt{2}-1]$

2 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}[\sqrt{2}-1]$

3 $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{d}}$

4 $\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}$

AP EAMCET (23.04.2018) Shift-1

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