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PHXII08:ELECTROMAGNETIC WAVES

368931 A condenser has two conducting plates of radius $10 c m$ separated by a distance of $5 m m$ . It is charged with a constant current of $0.15 A$ . The magnetic field at a point $2 c m$ from the axis in the gap is

1

3 \times 10^{- 8} T

2

3 \times 10^{- 6} T

3

6 \times 10^{- 8} T

4

1.5 \times 10^{- 6} T

Explanation:

$B = \frac{μ_{0} i r^{2}}{2 π R^{2}} = 6 \times 10^{- 8} T$

PHXII08:ELECTROMAGNETIC WAVES

368932 The diameter of the condenser plate is $4 c m$ . It is charged by an external current of $0.2 A$ . The maximum magnetic field induced in the gap

1

2 μ T

2

8 μ T

3

4 μ T

4

6 μ T

Explanation:

$B = \frac{μ_{0} i}{2 π r} = 2 μ T (r =$ radius of the plate $)$

PHXII08:ELECTROMAGNETIC WAVES

368933 An $A C$ peak voltage of $2 V$ having a frequency of $50 K H z$ is applied to a condenser of a capacity of $10 μ F$ . The maximum value of the magnetic field between the plates of the condenser if the radius of plate is $10 c m$ is

1

4 π μ T

2

40 π μ T

3

2 μ T

4

0.4 p μ

Explanation:

$B = \frac{μ_{o}}{2 π R} i = \frac{μ_{o}}{2 π R} \frac{d}{d t} [q_{o} \sin ω t]$
$B = \frac{μ_{o}}{2 π R} q_{o} ω \cos ω t$
$\begin{aligned} B = \frac{μ_{o}}{2 π R} q_{o} ω (q_{o} = C V_{o}) \\ B_{max} = 2 μ T \end{aligned}$

PHXII08:ELECTROMAGNETIC WAVES

368934 An electric field of $300 V / m$ is confined to a circular area $10 c m$ in diameter. If the field is increasing at the rate of $20 V / m - s$ , the magnitude of magnetic field at a point $15 c m$ from the center of the circle will be-

1

1.85 \times 10^{- 17} T

2

1.85 \times 10^{- 15} T

3

1.85 \times 10^{- 16} T

4

1.85 \times 10^{- 18} T

Explanation:

$B = \frac{μ_{0} ε_{0}}{2 π R} (\frac{π d^{2}}{4}) \frac{d E}{d t}$
$= \frac{2 \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 3.14 \times 0.01 \times 20}{4 \times 0.15}$
$= 1.85 \times 10^{- 18} T$

PHXII08:ELECTROMAGNETIC WAVES

368931 A condenser has two conducting plates of radius $10 c m$ separated by a distance of $5 m m$ . It is charged with a constant current of $0.15 A$ . The magnetic field at a point $2 c m$ from the axis in the gap is

1

3 \times 10^{- 8} T

2

3 \times 10^{- 6} T

3

6 \times 10^{- 8} T

4

1.5 \times 10^{- 6} T

Explanation:

$B = \frac{μ_{0} i r^{2}}{2 π R^{2}} = 6 \times 10^{- 8} T$

PHXII08:ELECTROMAGNETIC WAVES

368932 The diameter of the condenser plate is $4 c m$ . It is charged by an external current of $0.2 A$ . The maximum magnetic field induced in the gap

1

2 μ T

2

8 μ T

3

4 μ T

4

6 μ T

Explanation:

$B = \frac{μ_{0} i}{2 π r} = 2 μ T (r =$ radius of the plate $)$

PHXII08:ELECTROMAGNETIC WAVES

368933 An $A C$ peak voltage of $2 V$ having a frequency of $50 K H z$ is applied to a condenser of a capacity of $10 μ F$ . The maximum value of the magnetic field between the plates of the condenser if the radius of plate is $10 c m$ is

1

4 π μ T

2

40 π μ T

3

2 μ T

4

0.4 p μ

Explanation:

$B = \frac{μ_{o}}{2 π R} i = \frac{μ_{o}}{2 π R} \frac{d}{d t} [q_{o} \sin ω t]$
$B = \frac{μ_{o}}{2 π R} q_{o} ω \cos ω t$
$\begin{aligned} B = \frac{μ_{o}}{2 π R} q_{o} ω (q_{o} = C V_{o}) \\ B_{max} = 2 μ T \end{aligned}$

PHXII08:ELECTROMAGNETIC WAVES

368934 An electric field of $300 V / m$ is confined to a circular area $10 c m$ in diameter. If the field is increasing at the rate of $20 V / m - s$ , the magnitude of magnetic field at a point $15 c m$ from the center of the circle will be-

1

1.85 \times 10^{- 17} T

2

1.85 \times 10^{- 15} T

3

1.85 \times 10^{- 16} T

4

1.85 \times 10^{- 18} T

Explanation:

$B = \frac{μ_{0} ε_{0}}{2 π R} (\frac{π d^{2}}{4}) \frac{d E}{d t}$
$= \frac{2 \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 3.14 \times 0.01 \times 20}{4 \times 0.15}$
$= 1.85 \times 10^{- 18} T$

PHXII08:ELECTROMAGNETIC WAVES

368931 A condenser has two conducting plates of radius $10 c m$ separated by a distance of $5 m m$ . It is charged with a constant current of $0.15 A$ . The magnetic field at a point $2 c m$ from the axis in the gap is

1

3 \times 10^{- 8} T

2

3 \times 10^{- 6} T

3

6 \times 10^{- 8} T

4

1.5 \times 10^{- 6} T

Explanation:

$B = \frac{μ_{0} i r^{2}}{2 π R^{2}} = 6 \times 10^{- 8} T$

PHXII08:ELECTROMAGNETIC WAVES

368932 The diameter of the condenser plate is $4 c m$ . It is charged by an external current of $0.2 A$ . The maximum magnetic field induced in the gap

1

2 μ T

2

8 μ T

3

4 μ T

4

6 μ T

Explanation:

$B = \frac{μ_{0} i}{2 π r} = 2 μ T (r =$ radius of the plate $)$

PHXII08:ELECTROMAGNETIC WAVES

368933 An $A C$ peak voltage of $2 V$ having a frequency of $50 K H z$ is applied to a condenser of a capacity of $10 μ F$ . The maximum value of the magnetic field between the plates of the condenser if the radius of plate is $10 c m$ is

1

4 π μ T

2

40 π μ T

3

2 μ T

4

0.4 p μ

Explanation:

$B = \frac{μ_{o}}{2 π R} i = \frac{μ_{o}}{2 π R} \frac{d}{d t} [q_{o} \sin ω t]$
$B = \frac{μ_{o}}{2 π R} q_{o} ω \cos ω t$
$\begin{aligned} B = \frac{μ_{o}}{2 π R} q_{o} ω (q_{o} = C V_{o}) \\ B_{max} = 2 μ T \end{aligned}$

PHXII08:ELECTROMAGNETIC WAVES

368934 An electric field of $300 V / m$ is confined to a circular area $10 c m$ in diameter. If the field is increasing at the rate of $20 V / m - s$ , the magnitude of magnetic field at a point $15 c m$ from the center of the circle will be-

1

1.85 \times 10^{- 17} T

2

1.85 \times 10^{- 15} T

3

1.85 \times 10^{- 16} T

4

1.85 \times 10^{- 18} T

Explanation:

$B = \frac{μ_{0} ε_{0}}{2 π R} (\frac{π d^{2}}{4}) \frac{d E}{d t}$
$= \frac{2 \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 3.14 \times 0.01 \times 20}{4 \times 0.15}$
$= 1.85 \times 10^{- 18} T$

PHXII08:ELECTROMAGNETIC WAVES

368931 A condenser has two conducting plates of radius $10 c m$ separated by a distance of $5 m m$ . It is charged with a constant current of $0.15 A$ . The magnetic field at a point $2 c m$ from the axis in the gap is

1

3 \times 10^{- 8} T

2

3 \times 10^{- 6} T

3

6 \times 10^{- 8} T

4

1.5 \times 10^{- 6} T

Explanation:

$B = \frac{μ_{0} i r^{2}}{2 π R^{2}} = 6 \times 10^{- 8} T$

PHXII08:ELECTROMAGNETIC WAVES

368932 The diameter of the condenser plate is $4 c m$ . It is charged by an external current of $0.2 A$ . The maximum magnetic field induced in the gap

1

2 μ T

2

8 μ T

3

4 μ T

4

6 μ T

Explanation:

$B = \frac{μ_{0} i}{2 π r} = 2 μ T (r =$ radius of the plate $)$

PHXII08:ELECTROMAGNETIC WAVES

368933 An $A C$ peak voltage of $2 V$ having a frequency of $50 K H z$ is applied to a condenser of a capacity of $10 μ F$ . The maximum value of the magnetic field between the plates of the condenser if the radius of plate is $10 c m$ is

1

4 π μ T

2

40 π μ T

3

2 μ T

4

0.4 p μ

Explanation:

$B = \frac{μ_{o}}{2 π R} i = \frac{μ_{o}}{2 π R} \frac{d}{d t} [q_{o} \sin ω t]$
$B = \frac{μ_{o}}{2 π R} q_{o} ω \cos ω t$
$\begin{aligned} B = \frac{μ_{o}}{2 π R} q_{o} ω (q_{o} = C V_{o}) \\ B_{max} = 2 μ T \end{aligned}$

PHXII08:ELECTROMAGNETIC WAVES

368934 An electric field of $300 V / m$ is confined to a circular area $10 c m$ in diameter. If the field is increasing at the rate of $20 V / m - s$ , the magnitude of magnetic field at a point $15 c m$ from the center of the circle will be-

1

1.85 \times 10^{- 17} T

2

1.85 \times 10^{- 15} T

3

1.85 \times 10^{- 16} T

4

1.85 \times 10^{- 18} T

Explanation:

$B = \frac{μ_{0} ε_{0}}{2 π R} (\frac{π d^{2}}{4}) \frac{d E}{d t}$
$= \frac{2 \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 3.14 \times 0.01 \times 20}{4 \times 0.15}$
$= 1.85 \times 10^{- 18} T$

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