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Moving Charges & Magnetism

153273 The magnetic field at a point midway between two parallel long wires carrying currents in the same direction is $10 μ T$ . If the direction of the smaller current among them is reserved, the field becomes $30 μ T$ . The ratio of the larger to the smaller current in them is

1

3 : 1

2

2 : 1

3

4 : 1

4

3 : 2

5

3 : 4

Explanation:

B Given, $B_{1} = 10 μ T, B_{2} = 30 μ T$
Magnetic field at a distance of ' $r$ ' from long wire
$B = \frac{μ_{0} I}{2 π r}$
Case-I When current in same direction-
$B_{1} = \frac{μ_{0}}{2 π r} (I_{2} - I_{1})$
$∴ \frac{μ_{0}}{2 π r} (I_{2} - I_{1}) = 10 μ T$
Case-II When current is opposite direction-
$B_{2} = \frac{μ_{0}}{2 π r} (I_{2} + I_{1})$
$\frac{μ_{0}}{2 π r} (I_{2} + I_{1}) = 30 μ T$
Dividing equation (i) \& equation (ii), we get-
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{10}{30}$
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{1}{3}$
$3 I_{2} - 3 I_{1} = I_{2} + I_{1}$
$3 I_{2} - I_{2} - 3 I_{1} - I_{1} = 0$
$2 I_{2} - 4 I_{1} = 0$
$\frac{I_{2}}{I_{1}} = \frac{2}{1}$
The ratio of the larger to the smaller current in $I_{2} : I_{1} = 2 : 1$ .

Kerala CEE- 2013

Moving Charges & Magnetism

153274 Two long parallel wires carry currents $i_{1}$ and $i_{2}$ such that $i_{1} > i_{2}$ . When the currents are in the same direction, the magnetic field at a point midway between the wires is $6 \times 10^{- 6} T$ . If the direction of $i_{2}$ is reversed, the field becomes $3 \times$ $10^{- 5} T$ . The ratio $\frac{i_{1}}{i_{2}}$ is

1

\frac{1}{2}

2 2

3

\frac{2}{3}

4

\frac{3}{2}

5

\frac{1}{5}

Explanation:

D Given that, $B_{1} = 6 \times 10^{- 6} T$
$B_{2} = 3 \times 10^{- 5} T$
Magnetic field in long parallel wire
$B = \frac{μ_{0} i}{2 π r} (∵ i_{1} > i_{2})$
Case-I When currents are in same direction
$B_{1} = \frac{μ_{0}}{2 π r} (i_{1} - i_{2})$
Case-II When currents are in opposite direction
$B_{2} = \frac{μ_{0}}{2 π r} (i_{1} + i_{2})$
Dividing equation (i) by equation (ii), we get -
$\frac{B_{1}}{B_{2}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{6 \times 10^{- 6}}{3 \times 10^{- 5}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2 \times 10^{- 1}}{1} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2}{10} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{1}{5} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}} \Rightarrow i_{1} + i_{2} = 5 i_{1} - 5 i_{2}$
$5 i_{1} - 5 i_{2} - i_{1} - i_{2} = 0$
$4 i_{1} = 6 i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{6}{4} = \frac{3}{2}$
$\frac{i_{1}}{i_{2}} = \frac{3}{2}$

Kerala CEE - 2009

Moving Charges & Magnetism

153275 A circular coil of 5 turns and of $10 cm$ mean diameter is connected to a voltage source. If the resistance of the coil is $10 Ω$ , the voltage of the source so as to nullify the horizontal component of earth's magnetic field of 30 A turn $m^{1}$ at the centre of the coil should be

1

6 V

, plane of the coil normal to magnetic meridian

2

2 V

, plane of the coil normal to magnetic meridian

3

6 V

, plane of the coil along the magnetic meridian

4

2 V

, plane of the coil along the magnetic meridian

5

4 V

, plane of the coil normal to magnetic meridian

Explanation:

A Given that, number of circular Coil $= 5$ turn
Resistance of coil $= 10 Ω$
Radius of coil $= \frac{10}{2} = 5 cm$
$∵$ Magnetic field of 1 A turn $/ m = 4 π \times 10^{- 7}$
$∴$ Magnetic field of 30 A turn $/ m = 30 \times 4 π \times 10^{- 7}$
Magnetic field at centre of coil
$B = \frac{μ_{0}}{2 r} NI$
$B = \frac{μ_{0} N}{2 r} \frac{V}{R} (∵ I = \frac{V}{R})$
$V = \frac{2 rRB}{μ_{0} N}$
$V = \frac{2 (5 \times 10^{- 2}) \times 10 \times (30 \times 4 π \times 10^{- 7})}{(4 π \times 10^{- 7}) \times 5}$
$V = 600 \times 10^{- 2} = 6 V$
To multiply the horizontal component of magnetic field of earth, plane of the coil should be normal to magnetic field.

Kerala CEE 2007

Moving Charges & Magnetism

153276 A solenoid $600 mm$ long has 50 turns on it and is wound on an iron rod of $7.5 mm$ radius. Find the flux through the solenoid when the current in it is $3 A$ . The relative permeability of iron is 600 :

1

1.66 Wb

2

1.66 nWb

3

1.66 mWb

4

1.66 μ Wb

5

1.66 pWb

Explanation:

C Given,
Length of solenoid $(l) = 600 mm = 0.6 m$
Number of turns $= 50$ , current $= 3 A, μ_{r} = 600$
Radius of iron rod $= 7.5 mm = \frac{7.5}{1000} = 7.5 \times 10^{- 3} m$
$∵$ Magnetic flux $(ϕ) = \frac{μ_{r} μ_{0} N^{2} Ai}{l}$
$ϕ = \frac{600 \times 4 π \times 10^{- 7} (50)^{2} \times π {(7.5 \times 10^{- 3})}^{2} \times 3}{0.6}$
$ϕ = 0.001663 = 1.66 \times 10^{- 3} = 1.66 mWb$

Kerala CEE 2006

Moving Charges & Magnetism

153273 The magnetic field at a point midway between two parallel long wires carrying currents in the same direction is $10 μ T$ . If the direction of the smaller current among them is reserved, the field becomes $30 μ T$ . The ratio of the larger to the smaller current in them is

1

3 : 1

2

2 : 1

3

4 : 1

4

3 : 2

5

3 : 4

Explanation:

B Given, $B_{1} = 10 μ T, B_{2} = 30 μ T$
Magnetic field at a distance of ' $r$ ' from long wire
$B = \frac{μ_{0} I}{2 π r}$
Case-I When current in same direction-
$B_{1} = \frac{μ_{0}}{2 π r} (I_{2} - I_{1})$
$∴ \frac{μ_{0}}{2 π r} (I_{2} - I_{1}) = 10 μ T$
Case-II When current is opposite direction-
$B_{2} = \frac{μ_{0}}{2 π r} (I_{2} + I_{1})$
$\frac{μ_{0}}{2 π r} (I_{2} + I_{1}) = 30 μ T$
Dividing equation (i) \& equation (ii), we get-
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{10}{30}$
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{1}{3}$
$3 I_{2} - 3 I_{1} = I_{2} + I_{1}$
$3 I_{2} - I_{2} - 3 I_{1} - I_{1} = 0$
$2 I_{2} - 4 I_{1} = 0$
$\frac{I_{2}}{I_{1}} = \frac{2}{1}$
The ratio of the larger to the smaller current in $I_{2} : I_{1} = 2 : 1$ .

Kerala CEE- 2013

Moving Charges & Magnetism

153274 Two long parallel wires carry currents $i_{1}$ and $i_{2}$ such that $i_{1} > i_{2}$ . When the currents are in the same direction, the magnetic field at a point midway between the wires is $6 \times 10^{- 6} T$ . If the direction of $i_{2}$ is reversed, the field becomes $3 \times$ $10^{- 5} T$ . The ratio $\frac{i_{1}}{i_{2}}$ is

1

\frac{1}{2}

2 2

3

\frac{2}{3}

4

\frac{3}{2}

5

\frac{1}{5}

Explanation:

D Given that, $B_{1} = 6 \times 10^{- 6} T$
$B_{2} = 3 \times 10^{- 5} T$
Magnetic field in long parallel wire
$B = \frac{μ_{0} i}{2 π r} (∵ i_{1} > i_{2})$
Case-I When currents are in same direction
$B_{1} = \frac{μ_{0}}{2 π r} (i_{1} - i_{2})$
Case-II When currents are in opposite direction
$B_{2} = \frac{μ_{0}}{2 π r} (i_{1} + i_{2})$
Dividing equation (i) by equation (ii), we get -
$\frac{B_{1}}{B_{2}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{6 \times 10^{- 6}}{3 \times 10^{- 5}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2 \times 10^{- 1}}{1} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2}{10} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{1}{5} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}} \Rightarrow i_{1} + i_{2} = 5 i_{1} - 5 i_{2}$
$5 i_{1} - 5 i_{2} - i_{1} - i_{2} = 0$
$4 i_{1} = 6 i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{6}{4} = \frac{3}{2}$
$\frac{i_{1}}{i_{2}} = \frac{3}{2}$

Kerala CEE - 2009

Moving Charges & Magnetism

153275 A circular coil of 5 turns and of $10 cm$ mean diameter is connected to a voltage source. If the resistance of the coil is $10 Ω$ , the voltage of the source so as to nullify the horizontal component of earth's magnetic field of 30 A turn $m^{1}$ at the centre of the coil should be

1

6 V

, plane of the coil normal to magnetic meridian

2

2 V

, plane of the coil normal to magnetic meridian

3

6 V

, plane of the coil along the magnetic meridian

4

2 V

, plane of the coil along the magnetic meridian

5

4 V

, plane of the coil normal to magnetic meridian

Explanation:

A Given that, number of circular Coil $= 5$ turn
Resistance of coil $= 10 Ω$
Radius of coil $= \frac{10}{2} = 5 cm$
$∵$ Magnetic field of 1 A turn $/ m = 4 π \times 10^{- 7}$
$∴$ Magnetic field of 30 A turn $/ m = 30 \times 4 π \times 10^{- 7}$
Magnetic field at centre of coil
$B = \frac{μ_{0}}{2 r} NI$
$B = \frac{μ_{0} N}{2 r} \frac{V}{R} (∵ I = \frac{V}{R})$
$V = \frac{2 rRB}{μ_{0} N}$
$V = \frac{2 (5 \times 10^{- 2}) \times 10 \times (30 \times 4 π \times 10^{- 7})}{(4 π \times 10^{- 7}) \times 5}$
$V = 600 \times 10^{- 2} = 6 V$
To multiply the horizontal component of magnetic field of earth, plane of the coil should be normal to magnetic field.

Kerala CEE 2007

Moving Charges & Magnetism

153276 A solenoid $600 mm$ long has 50 turns on it and is wound on an iron rod of $7.5 mm$ radius. Find the flux through the solenoid when the current in it is $3 A$ . The relative permeability of iron is 600 :

1

1.66 Wb

2

1.66 nWb

3

1.66 mWb

4

1.66 μ Wb

5

1.66 pWb

Explanation:

C Given,
Length of solenoid $(l) = 600 mm = 0.6 m$
Number of turns $= 50$ , current $= 3 A, μ_{r} = 600$
Radius of iron rod $= 7.5 mm = \frac{7.5}{1000} = 7.5 \times 10^{- 3} m$
$∵$ Magnetic flux $(ϕ) = \frac{μ_{r} μ_{0} N^{2} Ai}{l}$
$ϕ = \frac{600 \times 4 π \times 10^{- 7} (50)^{2} \times π {(7.5 \times 10^{- 3})}^{2} \times 3}{0.6}$
$ϕ = 0.001663 = 1.66 \times 10^{- 3} = 1.66 mWb$

Kerala CEE 2006

Moving Charges & Magnetism

153273 The magnetic field at a point midway between two parallel long wires carrying currents in the same direction is $10 μ T$ . If the direction of the smaller current among them is reserved, the field becomes $30 μ T$ . The ratio of the larger to the smaller current in them is

1

3 : 1

2

2 : 1

3

4 : 1

4

3 : 2

5

3 : 4

Explanation:

B Given, $B_{1} = 10 μ T, B_{2} = 30 μ T$
Magnetic field at a distance of ' $r$ ' from long wire
$B = \frac{μ_{0} I}{2 π r}$
Case-I When current in same direction-
$B_{1} = \frac{μ_{0}}{2 π r} (I_{2} - I_{1})$
$∴ \frac{μ_{0}}{2 π r} (I_{2} - I_{1}) = 10 μ T$
Case-II When current is opposite direction-
$B_{2} = \frac{μ_{0}}{2 π r} (I_{2} + I_{1})$
$\frac{μ_{0}}{2 π r} (I_{2} + I_{1}) = 30 μ T$
Dividing equation (i) \& equation (ii), we get-
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{10}{30}$
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{1}{3}$
$3 I_{2} - 3 I_{1} = I_{2} + I_{1}$
$3 I_{2} - I_{2} - 3 I_{1} - I_{1} = 0$
$2 I_{2} - 4 I_{1} = 0$
$\frac{I_{2}}{I_{1}} = \frac{2}{1}$
The ratio of the larger to the smaller current in $I_{2} : I_{1} = 2 : 1$ .

Kerala CEE- 2013

Moving Charges & Magnetism

153274 Two long parallel wires carry currents $i_{1}$ and $i_{2}$ such that $i_{1} > i_{2}$ . When the currents are in the same direction, the magnetic field at a point midway between the wires is $6 \times 10^{- 6} T$ . If the direction of $i_{2}$ is reversed, the field becomes $3 \times$ $10^{- 5} T$ . The ratio $\frac{i_{1}}{i_{2}}$ is

1

\frac{1}{2}

2 2

3

\frac{2}{3}

4

\frac{3}{2}

5

\frac{1}{5}

Explanation:

D Given that, $B_{1} = 6 \times 10^{- 6} T$
$B_{2} = 3 \times 10^{- 5} T$
Magnetic field in long parallel wire
$B = \frac{μ_{0} i}{2 π r} (∵ i_{1} > i_{2})$
Case-I When currents are in same direction
$B_{1} = \frac{μ_{0}}{2 π r} (i_{1} - i_{2})$
Case-II When currents are in opposite direction
$B_{2} = \frac{μ_{0}}{2 π r} (i_{1} + i_{2})$
Dividing equation (i) by equation (ii), we get -
$\frac{B_{1}}{B_{2}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{6 \times 10^{- 6}}{3 \times 10^{- 5}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2 \times 10^{- 1}}{1} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2}{10} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{1}{5} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}} \Rightarrow i_{1} + i_{2} = 5 i_{1} - 5 i_{2}$
$5 i_{1} - 5 i_{2} - i_{1} - i_{2} = 0$
$4 i_{1} = 6 i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{6}{4} = \frac{3}{2}$
$\frac{i_{1}}{i_{2}} = \frac{3}{2}$

Kerala CEE - 2009

Moving Charges & Magnetism

153275 A circular coil of 5 turns and of $10 cm$ mean diameter is connected to a voltage source. If the resistance of the coil is $10 Ω$ , the voltage of the source so as to nullify the horizontal component of earth's magnetic field of 30 A turn $m^{1}$ at the centre of the coil should be

1

6 V

, plane of the coil normal to magnetic meridian

2

2 V

, plane of the coil normal to magnetic meridian

3

6 V

, plane of the coil along the magnetic meridian

4

2 V

, plane of the coil along the magnetic meridian

5

4 V

, plane of the coil normal to magnetic meridian

Explanation:

A Given that, number of circular Coil $= 5$ turn
Resistance of coil $= 10 Ω$
Radius of coil $= \frac{10}{2} = 5 cm$
$∵$ Magnetic field of 1 A turn $/ m = 4 π \times 10^{- 7}$
$∴$ Magnetic field of 30 A turn $/ m = 30 \times 4 π \times 10^{- 7}$
Magnetic field at centre of coil
$B = \frac{μ_{0}}{2 r} NI$
$B = \frac{μ_{0} N}{2 r} \frac{V}{R} (∵ I = \frac{V}{R})$
$V = \frac{2 rRB}{μ_{0} N}$
$V = \frac{2 (5 \times 10^{- 2}) \times 10 \times (30 \times 4 π \times 10^{- 7})}{(4 π \times 10^{- 7}) \times 5}$
$V = 600 \times 10^{- 2} = 6 V$
To multiply the horizontal component of magnetic field of earth, plane of the coil should be normal to magnetic field.

Kerala CEE 2007

Moving Charges & Magnetism

153276 A solenoid $600 mm$ long has 50 turns on it and is wound on an iron rod of $7.5 mm$ radius. Find the flux through the solenoid when the current in it is $3 A$ . The relative permeability of iron is 600 :

1

1.66 Wb

2

1.66 nWb

3

1.66 mWb

4

1.66 μ Wb

5

1.66 pWb

Explanation:

C Given,
Length of solenoid $(l) = 600 mm = 0.6 m$
Number of turns $= 50$ , current $= 3 A, μ_{r} = 600$
Radius of iron rod $= 7.5 mm = \frac{7.5}{1000} = 7.5 \times 10^{- 3} m$
$∵$ Magnetic flux $(ϕ) = \frac{μ_{r} μ_{0} N^{2} Ai}{l}$
$ϕ = \frac{600 \times 4 π \times 10^{- 7} (50)^{2} \times π {(7.5 \times 10^{- 3})}^{2} \times 3}{0.6}$
$ϕ = 0.001663 = 1.66 \times 10^{- 3} = 1.66 mWb$

Kerala CEE 2006

Moving Charges & Magnetism

153273 The magnetic field at a point midway between two parallel long wires carrying currents in the same direction is $10 μ T$ . If the direction of the smaller current among them is reserved, the field becomes $30 μ T$ . The ratio of the larger to the smaller current in them is

1

3 : 1

2

2 : 1

3

4 : 1

4

3 : 2

5

3 : 4

Explanation:

B Given, $B_{1} = 10 μ T, B_{2} = 30 μ T$
Magnetic field at a distance of ' $r$ ' from long wire
$B = \frac{μ_{0} I}{2 π r}$
Case-I When current in same direction-
$B_{1} = \frac{μ_{0}}{2 π r} (I_{2} - I_{1})$
$∴ \frac{μ_{0}}{2 π r} (I_{2} - I_{1}) = 10 μ T$
Case-II When current is opposite direction-
$B_{2} = \frac{μ_{0}}{2 π r} (I_{2} + I_{1})$
$\frac{μ_{0}}{2 π r} (I_{2} + I_{1}) = 30 μ T$
Dividing equation (i) \& equation (ii), we get-
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{10}{30}$
$\frac{I_{2} - I_{1}}{I_{2} + I_{1}} = \frac{1}{3}$
$3 I_{2} - 3 I_{1} = I_{2} + I_{1}$
$3 I_{2} - I_{2} - 3 I_{1} - I_{1} = 0$
$2 I_{2} - 4 I_{1} = 0$
$\frac{I_{2}}{I_{1}} = \frac{2}{1}$
The ratio of the larger to the smaller current in $I_{2} : I_{1} = 2 : 1$ .

Kerala CEE- 2013

Moving Charges & Magnetism

153274 Two long parallel wires carry currents $i_{1}$ and $i_{2}$ such that $i_{1} > i_{2}$ . When the currents are in the same direction, the magnetic field at a point midway between the wires is $6 \times 10^{- 6} T$ . If the direction of $i_{2}$ is reversed, the field becomes $3 \times$ $10^{- 5} T$ . The ratio $\frac{i_{1}}{i_{2}}$ is

1

\frac{1}{2}

2 2

3

\frac{2}{3}

4

\frac{3}{2}

5

\frac{1}{5}

Explanation:

D Given that, $B_{1} = 6 \times 10^{- 6} T$
$B_{2} = 3 \times 10^{- 5} T$
Magnetic field in long parallel wire
$B = \frac{μ_{0} i}{2 π r} (∵ i_{1} > i_{2})$
Case-I When currents are in same direction
$B_{1} = \frac{μ_{0}}{2 π r} (i_{1} - i_{2})$
Case-II When currents are in opposite direction
$B_{2} = \frac{μ_{0}}{2 π r} (i_{1} + i_{2})$
Dividing equation (i) by equation (ii), we get -
$\frac{B_{1}}{B_{2}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{6 \times 10^{- 6}}{3 \times 10^{- 5}} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2 \times 10^{- 1}}{1} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{2}{10} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}}$
$\frac{1}{5} = \frac{i_{1} - i_{2}}{i_{1} + i_{2}} \Rightarrow i_{1} + i_{2} = 5 i_{1} - 5 i_{2}$
$5 i_{1} - 5 i_{2} - i_{1} - i_{2} = 0$
$4 i_{1} = 6 i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{6}{4} = \frac{3}{2}$
$\frac{i_{1}}{i_{2}} = \frac{3}{2}$

Kerala CEE - 2009

Moving Charges & Magnetism

153275 A circular coil of 5 turns and of $10 cm$ mean diameter is connected to a voltage source. If the resistance of the coil is $10 Ω$ , the voltage of the source so as to nullify the horizontal component of earth's magnetic field of 30 A turn $m^{1}$ at the centre of the coil should be

1

6 V

, plane of the coil normal to magnetic meridian

2

2 V

, plane of the coil normal to magnetic meridian

3

6 V

, plane of the coil along the magnetic meridian

4

2 V

, plane of the coil along the magnetic meridian

5

4 V

, plane of the coil normal to magnetic meridian

Explanation:

A Given that, number of circular Coil $= 5$ turn
Resistance of coil $= 10 Ω$
Radius of coil $= \frac{10}{2} = 5 cm$
$∵$ Magnetic field of 1 A turn $/ m = 4 π \times 10^{- 7}$
$∴$ Magnetic field of 30 A turn $/ m = 30 \times 4 π \times 10^{- 7}$
Magnetic field at centre of coil
$B = \frac{μ_{0}}{2 r} NI$
$B = \frac{μ_{0} N}{2 r} \frac{V}{R} (∵ I = \frac{V}{R})$
$V = \frac{2 rRB}{μ_{0} N}$
$V = \frac{2 (5 \times 10^{- 2}) \times 10 \times (30 \times 4 π \times 10^{- 7})}{(4 π \times 10^{- 7}) \times 5}$
$V = 600 \times 10^{- 2} = 6 V$
To multiply the horizontal component of magnetic field of earth, plane of the coil should be normal to magnetic field.

Kerala CEE 2007

Moving Charges & Magnetism

153276 A solenoid $600 mm$ long has 50 turns on it and is wound on an iron rod of $7.5 mm$ radius. Find the flux through the solenoid when the current in it is $3 A$ . The relative permeability of iron is 600 :

1

1.66 Wb

2

1.66 nWb

3

1.66 mWb

4

1.66 μ Wb

5

1.66 pWb

Explanation:

C Given,
Length of solenoid $(l) = 600 mm = 0.6 m$
Number of turns $= 50$ , current $= 3 A, μ_{r} = 600$
Radius of iron rod $= 7.5 mm = \frac{7.5}{1000} = 7.5 \times 10^{- 3} m$
$∵$ Magnetic flux $(ϕ) = \frac{μ_{r} μ_{0} N^{2} Ai}{l}$
$ϕ = \frac{600 \times 4 π \times 10^{- 7} (50)^{2} \times π {(7.5 \times 10^{- 3})}^{2} \times 3}{0.6}$
$ϕ = 0.001663 = 1.66 \times 10^{- 3} = 1.66 mWb$

Kerala CEE 2006

Question Bank Series

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Question Bank Series

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