QBManager

Moving Charges & Magnetism

153200 Two wire loops $P Q R S P$ formed by joining two semicircle wires of radii $R_{1}$ and $R_{2}$ carrying a current $I$ as shown in figure. The magnetic field at centre $C$ is

1

\frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

2

\frac{μ_{0} I}{4} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

3

\frac{μ_{0} I}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

4

\frac{μ_{0} I}{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

Explanation:

A Magnetic field at the centre of coil
$B_{C} = \frac{μ_{0} IN}{2 r}$
For semi circular $N = 0.5$
$B_{1} = \frac{μ_{0} I (0.5)}{2 R_{1}} = \frac{μ_{0} I}{4 R_{1}}$
$B_{2} = \frac{μ_{0} I (0.5)}{2 R_{2}} = \frac{μ_{0} I}{4 R_{2}}$
So, Net Magnetic field
$B_{Net} = B_{1} - B_{2}$
$= \frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

CG PET -2019

Moving Charges & Magnetism

153202 A magnetic field of $5 \times 10^{- 5} T$ is produced at a perpendicular distance of $0.2 m$ from a long straight wire carrying electric current. If the permeability of free space is $4 π \times 10^{- 7} T m / A$ . The current passing through the wire in $A$ is

1 45

2 40

3 50

4 30

Explanation:

C Given,
$B = 5 \times 10^{- 5} T$
$r = 0.2 m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field produced by a wire
$B = \frac{μ_{0}}{2 π} \frac{I}{r}$
$5 \times 10^{- 5} = \frac{4 π \times 10^{- 7}}{2 π} \times \frac{I}{0.2}$
$I = \frac{5 \times 10^{- 5} \times 0.2 \times 2 π}{4 π \times 10^{- 7}}$
$I = 0.5 \times 10^{2} A$
$I = 50 A$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153203 A long wire carries a current of $18 A$ kept along the axis of a long solenoid of radius $1 cm$ . The field due to the solenoid is $8.0 \times 10^{- 3} T$ . The magnitude of the resultant field at a point 0.6 $mm$ from the solenoid axis is (Assume, $μ_{0} = 4 π$ $\times 10^{- 7} Tm / A)$

1

6 \times 10^{- 3} T

2

6 \times 10^{- 4} T

3

2 \sqrt{7} \times 10^{- 3} T

4

10 \times 10^{- 3} T

Explanation:

D Given,
$I = 18 A$
$R = 1 cm = 10^{- 2} m$
$B_{1} = 8 \times 10^{- 3} T$
$r = 0.6 mm = 0.6 \times 10^{- 3} m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field due to current carrying wire
$B_{2} = \frac{μ_{0}}{2 π} \times \frac{I}{r}$
$= \frac{4 π \times 10^{- 7}}{2 π} \times \frac{18}{0.6 \times 10^{- 3}} = 6 \times 10^{- 3} T$
Resultant field magnitude
$B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(8 \times 10^{- 3})}^{2} + {(6 \times 10^{- 3})}^{2}}$
$B = 10 \times 10^{- 3} T$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153204 A dielectric circular dise of radius $R$ carries a uniform surface charge density $σ$ . If it rotates about its axis with angular velocity $ω$ , the magnetic field at the centre of disc is :

1

\frac{μ_{0} σ ω R^{2}}{2 π}

2

\frac{μ_{0} σ ω R}{2}

3

\frac{μ_{0} σ ω R^{2}}{4}

4

\frac{μ_{0} σ ω R^{2}}{2 \sqrt{2}}

Explanation:

B Charge on the element $dq = σ (2 π r) dr$
Current di associated with rotating charge $dq$ is
$di = \frac{dq}{T} = \frac{dq ω}{2 π} (∵ T = \frac{2 π}{ω})$

From equation (i) and (ii) we get
$di = σ ω rdr$
As, The magnetic field $dB$ at centre due to element.
$dB = \frac{μ_{0} di}{2 r} = \frac{μ_{0} σ ω rdr}{2 r}$
$dB = \frac{μ_{0} σ ω}{2} dr$
$\int dB = \frac{μ_{0} σ ω}{2} \int_{0}^{R} dr$
$B_{Net} = \frac{μ_{0} σ ω R}{2}$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153205 A long straight wire carrying current $16 A$ is bent at $90^{\circ}$ such that half of the wire lies along the positive $x$ -axis and other half lies along the positive $y$ -axis. What is the magnitude of the magnetic field at the point, $r = (- 2 \hat{i} + 0 \hat{j}) m m$ ?
$(Assume, \frac{μ_{0}}{4 π} = 10^{- 7} H m^{- 1})$

1

1.2 mT

2

0.8 mT

3

3.2 mT

4

1.6 mT

Explanation:

B Given, $I = 16 A, r = 2 mm = 2 \times 10^{- 3} m$ According to the question a long straight wire is bent

$B = \frac{μ_{0}}{4 π} \frac{I}{r}$
Magnetic field at point $P$ , due to wire along $x$ -axis is zero. Magnetic field due to wire along $y$ -axis at point $p$ is given by-
$B = \frac{μ_{0}}{4 π} \times \frac{16}{2 \times 10^{- 3}}$
$B = 10^{- 7} \times 8 \times 10^{3} = 0.8 \times 10^{- 3} T$
$B = 0.8 mT$

TS- EAMCET-03.05.2019

Moving Charges & Magnetism

153200 Two wire loops $P Q R S P$ formed by joining two semicircle wires of radii $R_{1}$ and $R_{2}$ carrying a current $I$ as shown in figure. The magnetic field at centre $C$ is

1

\frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

2

\frac{μ_{0} I}{4} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

3

\frac{μ_{0} I}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

4

\frac{μ_{0} I}{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

Explanation:

A Magnetic field at the centre of coil
$B_{C} = \frac{μ_{0} IN}{2 r}$
For semi circular $N = 0.5$
$B_{1} = \frac{μ_{0} I (0.5)}{2 R_{1}} = \frac{μ_{0} I}{4 R_{1}}$
$B_{2} = \frac{μ_{0} I (0.5)}{2 R_{2}} = \frac{μ_{0} I}{4 R_{2}}$
So, Net Magnetic field
$B_{Net} = B_{1} - B_{2}$
$= \frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

CG PET -2019

Moving Charges & Magnetism

153202 A magnetic field of $5 \times 10^{- 5} T$ is produced at a perpendicular distance of $0.2 m$ from a long straight wire carrying electric current. If the permeability of free space is $4 π \times 10^{- 7} T m / A$ . The current passing through the wire in $A$ is

1 45

2 40

3 50

4 30

Explanation:

C Given,
$B = 5 \times 10^{- 5} T$
$r = 0.2 m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field produced by a wire
$B = \frac{μ_{0}}{2 π} \frac{I}{r}$
$5 \times 10^{- 5} = \frac{4 π \times 10^{- 7}}{2 π} \times \frac{I}{0.2}$
$I = \frac{5 \times 10^{- 5} \times 0.2 \times 2 π}{4 π \times 10^{- 7}}$
$I = 0.5 \times 10^{2} A$
$I = 50 A$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153203 A long wire carries a current of $18 A$ kept along the axis of a long solenoid of radius $1 cm$ . The field due to the solenoid is $8.0 \times 10^{- 3} T$ . The magnitude of the resultant field at a point 0.6 $mm$ from the solenoid axis is (Assume, $μ_{0} = 4 π$ $\times 10^{- 7} Tm / A)$

1

6 \times 10^{- 3} T

2

6 \times 10^{- 4} T

3

2 \sqrt{7} \times 10^{- 3} T

4

10 \times 10^{- 3} T

Explanation:

D Given,
$I = 18 A$
$R = 1 cm = 10^{- 2} m$
$B_{1} = 8 \times 10^{- 3} T$
$r = 0.6 mm = 0.6 \times 10^{- 3} m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field due to current carrying wire
$B_{2} = \frac{μ_{0}}{2 π} \times \frac{I}{r}$
$= \frac{4 π \times 10^{- 7}}{2 π} \times \frac{18}{0.6 \times 10^{- 3}} = 6 \times 10^{- 3} T$
Resultant field magnitude
$B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(8 \times 10^{- 3})}^{2} + {(6 \times 10^{- 3})}^{2}}$
$B = 10 \times 10^{- 3} T$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153204 A dielectric circular dise of radius $R$ carries a uniform surface charge density $σ$ . If it rotates about its axis with angular velocity $ω$ , the magnetic field at the centre of disc is :

1

\frac{μ_{0} σ ω R^{2}}{2 π}

2

\frac{μ_{0} σ ω R}{2}

3

\frac{μ_{0} σ ω R^{2}}{4}

4

\frac{μ_{0} σ ω R^{2}}{2 \sqrt{2}}

Explanation:

B Charge on the element $dq = σ (2 π r) dr$
Current di associated with rotating charge $dq$ is
$di = \frac{dq}{T} = \frac{dq ω}{2 π} (∵ T = \frac{2 π}{ω})$

From equation (i) and (ii) we get
$di = σ ω rdr$
As, The magnetic field $dB$ at centre due to element.
$dB = \frac{μ_{0} di}{2 r} = \frac{μ_{0} σ ω rdr}{2 r}$
$dB = \frac{μ_{0} σ ω}{2} dr$
$\int dB = \frac{μ_{0} σ ω}{2} \int_{0}^{R} dr$
$B_{Net} = \frac{μ_{0} σ ω R}{2}$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153205 A long straight wire carrying current $16 A$ is bent at $90^{\circ}$ such that half of the wire lies along the positive $x$ -axis and other half lies along the positive $y$ -axis. What is the magnitude of the magnetic field at the point, $r = (- 2 \hat{i} + 0 \hat{j}) m m$ ?
$(Assume, \frac{μ_{0}}{4 π} = 10^{- 7} H m^{- 1})$

1

1.2 mT

2

0.8 mT

3

3.2 mT

4

1.6 mT

Explanation:

B Given, $I = 16 A, r = 2 mm = 2 \times 10^{- 3} m$ According to the question a long straight wire is bent

$B = \frac{μ_{0}}{4 π} \frac{I}{r}$
Magnetic field at point $P$ , due to wire along $x$ -axis is zero. Magnetic field due to wire along $y$ -axis at point $p$ is given by-
$B = \frac{μ_{0}}{4 π} \times \frac{16}{2 \times 10^{- 3}}$
$B = 10^{- 7} \times 8 \times 10^{3} = 0.8 \times 10^{- 3} T$
$B = 0.8 mT$

TS- EAMCET-03.05.2019

Moving Charges & Magnetism

153200 Two wire loops $P Q R S P$ formed by joining two semicircle wires of radii $R_{1}$ and $R_{2}$ carrying a current $I$ as shown in figure. The magnetic field at centre $C$ is

1

\frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

2

\frac{μ_{0} I}{4} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

3

\frac{μ_{0} I}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

4

\frac{μ_{0} I}{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

Explanation:

A Magnetic field at the centre of coil
$B_{C} = \frac{μ_{0} IN}{2 r}$
For semi circular $N = 0.5$
$B_{1} = \frac{μ_{0} I (0.5)}{2 R_{1}} = \frac{μ_{0} I}{4 R_{1}}$
$B_{2} = \frac{μ_{0} I (0.5)}{2 R_{2}} = \frac{μ_{0} I}{4 R_{2}}$
So, Net Magnetic field
$B_{Net} = B_{1} - B_{2}$
$= \frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

CG PET -2019

Moving Charges & Magnetism

153202 A magnetic field of $5 \times 10^{- 5} T$ is produced at a perpendicular distance of $0.2 m$ from a long straight wire carrying electric current. If the permeability of free space is $4 π \times 10^{- 7} T m / A$ . The current passing through the wire in $A$ is

1 45

2 40

3 50

4 30

Explanation:

C Given,
$B = 5 \times 10^{- 5} T$
$r = 0.2 m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field produced by a wire
$B = \frac{μ_{0}}{2 π} \frac{I}{r}$
$5 \times 10^{- 5} = \frac{4 π \times 10^{- 7}}{2 π} \times \frac{I}{0.2}$
$I = \frac{5 \times 10^{- 5} \times 0.2 \times 2 π}{4 π \times 10^{- 7}}$
$I = 0.5 \times 10^{2} A$
$I = 50 A$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153203 A long wire carries a current of $18 A$ kept along the axis of a long solenoid of radius $1 cm$ . The field due to the solenoid is $8.0 \times 10^{- 3} T$ . The magnitude of the resultant field at a point 0.6 $mm$ from the solenoid axis is (Assume, $μ_{0} = 4 π$ $\times 10^{- 7} Tm / A)$

1

6 \times 10^{- 3} T

2

6 \times 10^{- 4} T

3

2 \sqrt{7} \times 10^{- 3} T

4

10 \times 10^{- 3} T

Explanation:

D Given,
$I = 18 A$
$R = 1 cm = 10^{- 2} m$
$B_{1} = 8 \times 10^{- 3} T$
$r = 0.6 mm = 0.6 \times 10^{- 3} m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field due to current carrying wire
$B_{2} = \frac{μ_{0}}{2 π} \times \frac{I}{r}$
$= \frac{4 π \times 10^{- 7}}{2 π} \times \frac{18}{0.6 \times 10^{- 3}} = 6 \times 10^{- 3} T$
Resultant field magnitude
$B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(8 \times 10^{- 3})}^{2} + {(6 \times 10^{- 3})}^{2}}$
$B = 10 \times 10^{- 3} T$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153204 A dielectric circular dise of radius $R$ carries a uniform surface charge density $σ$ . If it rotates about its axis with angular velocity $ω$ , the magnetic field at the centre of disc is :

1

\frac{μ_{0} σ ω R^{2}}{2 π}

2

\frac{μ_{0} σ ω R}{2}

3

\frac{μ_{0} σ ω R^{2}}{4}

4

\frac{μ_{0} σ ω R^{2}}{2 \sqrt{2}}

Explanation:

B Charge on the element $dq = σ (2 π r) dr$
Current di associated with rotating charge $dq$ is
$di = \frac{dq}{T} = \frac{dq ω}{2 π} (∵ T = \frac{2 π}{ω})$

From equation (i) and (ii) we get
$di = σ ω rdr$
As, The magnetic field $dB$ at centre due to element.
$dB = \frac{μ_{0} di}{2 r} = \frac{μ_{0} σ ω rdr}{2 r}$
$dB = \frac{μ_{0} σ ω}{2} dr$
$\int dB = \frac{μ_{0} σ ω}{2} \int_{0}^{R} dr$
$B_{Net} = \frac{μ_{0} σ ω R}{2}$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153205 A long straight wire carrying current $16 A$ is bent at $90^{\circ}$ such that half of the wire lies along the positive $x$ -axis and other half lies along the positive $y$ -axis. What is the magnitude of the magnetic field at the point, $r = (- 2 \hat{i} + 0 \hat{j}) m m$ ?
$(Assume, \frac{μ_{0}}{4 π} = 10^{- 7} H m^{- 1})$

1

1.2 mT

2

0.8 mT

3

3.2 mT

4

1.6 mT

Explanation:

B Given, $I = 16 A, r = 2 mm = 2 \times 10^{- 3} m$ According to the question a long straight wire is bent

$B = \frac{μ_{0}}{4 π} \frac{I}{r}$
Magnetic field at point $P$ , due to wire along $x$ -axis is zero. Magnetic field due to wire along $y$ -axis at point $p$ is given by-
$B = \frac{μ_{0}}{4 π} \times \frac{16}{2 \times 10^{- 3}}$
$B = 10^{- 7} \times 8 \times 10^{3} = 0.8 \times 10^{- 3} T$
$B = 0.8 mT$

TS- EAMCET-03.05.2019

Moving Charges & Magnetism

153200 Two wire loops $P Q R S P$ formed by joining two semicircle wires of radii $R_{1}$ and $R_{2}$ carrying a current $I$ as shown in figure. The magnetic field at centre $C$ is

1

\frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

2

\frac{μ_{0} I}{4} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

3

\frac{μ_{0} I}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

4

\frac{μ_{0} I}{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

Explanation:

A Magnetic field at the centre of coil
$B_{C} = \frac{μ_{0} IN}{2 r}$
For semi circular $N = 0.5$
$B_{1} = \frac{μ_{0} I (0.5)}{2 R_{1}} = \frac{μ_{0} I}{4 R_{1}}$
$B_{2} = \frac{μ_{0} I (0.5)}{2 R_{2}} = \frac{μ_{0} I}{4 R_{2}}$
So, Net Magnetic field
$B_{Net} = B_{1} - B_{2}$
$= \frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

CG PET -2019

Moving Charges & Magnetism

153202 A magnetic field of $5 \times 10^{- 5} T$ is produced at a perpendicular distance of $0.2 m$ from a long straight wire carrying electric current. If the permeability of free space is $4 π \times 10^{- 7} T m / A$ . The current passing through the wire in $A$ is

1 45

2 40

3 50

4 30

Explanation:

C Given,
$B = 5 \times 10^{- 5} T$
$r = 0.2 m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field produced by a wire
$B = \frac{μ_{0}}{2 π} \frac{I}{r}$
$5 \times 10^{- 5} = \frac{4 π \times 10^{- 7}}{2 π} \times \frac{I}{0.2}$
$I = \frac{5 \times 10^{- 5} \times 0.2 \times 2 π}{4 π \times 10^{- 7}}$
$I = 0.5 \times 10^{2} A$
$I = 50 A$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153203 A long wire carries a current of $18 A$ kept along the axis of a long solenoid of radius $1 cm$ . The field due to the solenoid is $8.0 \times 10^{- 3} T$ . The magnitude of the resultant field at a point 0.6 $mm$ from the solenoid axis is (Assume, $μ_{0} = 4 π$ $\times 10^{- 7} Tm / A)$

1

6 \times 10^{- 3} T

2

6 \times 10^{- 4} T

3

2 \sqrt{7} \times 10^{- 3} T

4

10 \times 10^{- 3} T

Explanation:

D Given,
$I = 18 A$
$R = 1 cm = 10^{- 2} m$
$B_{1} = 8 \times 10^{- 3} T$
$r = 0.6 mm = 0.6 \times 10^{- 3} m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field due to current carrying wire
$B_{2} = \frac{μ_{0}}{2 π} \times \frac{I}{r}$
$= \frac{4 π \times 10^{- 7}}{2 π} \times \frac{18}{0.6 \times 10^{- 3}} = 6 \times 10^{- 3} T$
Resultant field magnitude
$B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(8 \times 10^{- 3})}^{2} + {(6 \times 10^{- 3})}^{2}}$
$B = 10 \times 10^{- 3} T$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153204 A dielectric circular dise of radius $R$ carries a uniform surface charge density $σ$ . If it rotates about its axis with angular velocity $ω$ , the magnetic field at the centre of disc is :

1

\frac{μ_{0} σ ω R^{2}}{2 π}

2

\frac{μ_{0} σ ω R}{2}

3

\frac{μ_{0} σ ω R^{2}}{4}

4

\frac{μ_{0} σ ω R^{2}}{2 \sqrt{2}}

Explanation:

B Charge on the element $dq = σ (2 π r) dr$
Current di associated with rotating charge $dq$ is
$di = \frac{dq}{T} = \frac{dq ω}{2 π} (∵ T = \frac{2 π}{ω})$

From equation (i) and (ii) we get
$di = σ ω rdr$
As, The magnetic field $dB$ at centre due to element.
$dB = \frac{μ_{0} di}{2 r} = \frac{μ_{0} σ ω rdr}{2 r}$
$dB = \frac{μ_{0} σ ω}{2} dr$
$\int dB = \frac{μ_{0} σ ω}{2} \int_{0}^{R} dr$
$B_{Net} = \frac{μ_{0} σ ω R}{2}$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153205 A long straight wire carrying current $16 A$ is bent at $90^{\circ}$ such that half of the wire lies along the positive $x$ -axis and other half lies along the positive $y$ -axis. What is the magnitude of the magnetic field at the point, $r = (- 2 \hat{i} + 0 \hat{j}) m m$ ?
$(Assume, \frac{μ_{0}}{4 π} = 10^{- 7} H m^{- 1})$

1

1.2 mT

2

0.8 mT

3

3.2 mT

4

1.6 mT

Explanation:

B Given, $I = 16 A, r = 2 mm = 2 \times 10^{- 3} m$ According to the question a long straight wire is bent

$B = \frac{μ_{0}}{4 π} \frac{I}{r}$
Magnetic field at point $P$ , due to wire along $x$ -axis is zero. Magnetic field due to wire along $y$ -axis at point $p$ is given by-
$B = \frac{μ_{0}}{4 π} \times \frac{16}{2 \times 10^{- 3}}$
$B = 10^{- 7} \times 8 \times 10^{3} = 0.8 \times 10^{- 3} T$
$B = 0.8 mT$

TS- EAMCET-03.05.2019

Moving Charges & Magnetism

153200 Two wire loops $P Q R S P$ formed by joining two semicircle wires of radii $R_{1}$ and $R_{2}$ carrying a current $I$ as shown in figure. The magnetic field at centre $C$ is

1

\frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

2

\frac{μ_{0} I}{4} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

3

\frac{μ_{0} I}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

4

\frac{μ_{0} I}{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

Explanation:

A Magnetic field at the centre of coil
$B_{C} = \frac{μ_{0} IN}{2 r}$
For semi circular $N = 0.5$
$B_{1} = \frac{μ_{0} I (0.5)}{2 R_{1}} = \frac{μ_{0} I}{4 R_{1}}$
$B_{2} = \frac{μ_{0} I (0.5)}{2 R_{2}} = \frac{μ_{0} I}{4 R_{2}}$
So, Net Magnetic field
$B_{Net} = B_{1} - B_{2}$
$= \frac{μ_{0} I}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

CG PET -2019

Moving Charges & Magnetism

153202 A magnetic field of $5 \times 10^{- 5} T$ is produced at a perpendicular distance of $0.2 m$ from a long straight wire carrying electric current. If the permeability of free space is $4 π \times 10^{- 7} T m / A$ . The current passing through the wire in $A$ is

1 45

2 40

3 50

4 30

Explanation:

C Given,
$B = 5 \times 10^{- 5} T$
$r = 0.2 m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field produced by a wire
$B = \frac{μ_{0}}{2 π} \frac{I}{r}$
$5 \times 10^{- 5} = \frac{4 π \times 10^{- 7}}{2 π} \times \frac{I}{0.2}$
$I = \frac{5 \times 10^{- 5} \times 0.2 \times 2 π}{4 π \times 10^{- 7}}$
$I = 0.5 \times 10^{2} A$
$I = 50 A$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153203 A long wire carries a current of $18 A$ kept along the axis of a long solenoid of radius $1 cm$ . The field due to the solenoid is $8.0 \times 10^{- 3} T$ . The magnitude of the resultant field at a point 0.6 $mm$ from the solenoid axis is (Assume, $μ_{0} = 4 π$ $\times 10^{- 7} Tm / A)$

1

6 \times 10^{- 3} T

2

6 \times 10^{- 4} T

3

2 \sqrt{7} \times 10^{- 3} T

4

10 \times 10^{- 3} T

Explanation:

D Given,
$I = 18 A$
$R = 1 cm = 10^{- 2} m$
$B_{1} = 8 \times 10^{- 3} T$
$r = 0.6 mm = 0.6 \times 10^{- 3} m$
$μ_{0} = 4 π \times 10^{- 7} Tm / A$
Magnetic field due to current carrying wire
$B_{2} = \frac{μ_{0}}{2 π} \times \frac{I}{r}$
$= \frac{4 π \times 10^{- 7}}{2 π} \times \frac{18}{0.6 \times 10^{- 3}} = 6 \times 10^{- 3} T$
Resultant field magnitude
$B = \sqrt{B_{1}^{2} + B_{2}^{2}}$
$= \sqrt{{(8 \times 10^{- 3})}^{2} + {(6 \times 10^{- 3})}^{2}}$
$B = 10 \times 10^{- 3} T$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153204 A dielectric circular dise of radius $R$ carries a uniform surface charge density $σ$ . If it rotates about its axis with angular velocity $ω$ , the magnetic field at the centre of disc is :

1

\frac{μ_{0} σ ω R^{2}}{2 π}

2

\frac{μ_{0} σ ω R}{2}

3

\frac{μ_{0} σ ω R^{2}}{4}

4

\frac{μ_{0} σ ω R^{2}}{2 \sqrt{2}}

Explanation:

B Charge on the element $dq = σ (2 π r) dr$
Current di associated with rotating charge $dq$ is
$di = \frac{dq}{T} = \frac{dq ω}{2 π} (∵ T = \frac{2 π}{ω})$

From equation (i) and (ii) we get
$di = σ ω rdr$
As, The magnetic field $dB$ at centre due to element.
$dB = \frac{μ_{0} di}{2 r} = \frac{μ_{0} σ ω rdr}{2 r}$
$dB = \frac{μ_{0} σ ω}{2} dr$
$\int dB = \frac{μ_{0} σ ω}{2} \int_{0}^{R} dr$
$B_{Net} = \frac{μ_{0} σ ω R}{2}$

TS- EAMCET-04.05.2019

Moving Charges & Magnetism

153205 A long straight wire carrying current $16 A$ is bent at $90^{\circ}$ such that half of the wire lies along the positive $x$ -axis and other half lies along the positive $y$ -axis. What is the magnitude of the magnetic field at the point, $r = (- 2 \hat{i} + 0 \hat{j}) m m$ ?
$(Assume, \frac{μ_{0}}{4 π} = 10^{- 7} H m^{- 1})$

1

1.2 mT

2

0.8 mT

3

3.2 mT

4

1.6 mT

Explanation:

B Given, $I = 16 A, r = 2 mm = 2 \times 10^{- 3} m$ According to the question a long straight wire is bent

$B = \frac{μ_{0}}{4 π} \frac{I}{r}$
Magnetic field at point $P$ , due to wire along $x$ -axis is zero. Magnetic field due to wire along $y$ -axis at point $p$ is given by-
$B = \frac{μ_{0}}{4 π} \times \frac{16}{2 \times 10^{- 3}}$
$B = 10^{- 7} \times 8 \times 10^{3} = 0.8 \times 10^{- 3} T$
$B = 0.8 mT$

TS- EAMCET-03.05.2019