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PHXII04:MOVING CHARGES AND MAGNETISM

362985 The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 %$ . The percentage change in voltage sensitivity of the galvanometer will be

1

0 %

2

50 %

3

75 %

4

100 %

Explanation:

Sensitivity of a moving coil galvanometer
$= \frac{N A B}{k}$
Deflection in current is increased but current sensitivity is independent of voltage sensitivity. Hence, percentage change in voltage sensitivity of the galvanometer is zero.
So, correct option is (1).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362986 For a moving coil galvanometer, the deflection in the coil is $0.05 r a d$ when a current of $10 m A$ is passed through it. If the torsional constant of suspension wire is $4.0 \times 10^{- 5} N m r a d^{- 1}$ , the magnetic field is $0.01 T$ and the number of turns in the coils is 200 , the area of each turn (in $c m^{2}$ ) is

1 2.0

2 1.0

3 1.5

4 0.5

Explanation:

$θ = 0.05 r a d$ , Current, $i = 10 m A$ ;
Torsional constant, $k = 4 \times 10^{- 5} N m / r a d$
$B = 0.01 T, N = 200$
Let area is $A$ .
At equilibrium
$τ = k θ$
$N i A B = k θ$
$A = \frac{k θ}{N i B} = \frac{4 \times 10^{- 5} \times 0.05}{200 \times 10 \times 10^{- 3} \times 0.01} m^{2}$
$A = 1 c m^{2}$
So, correction is (2).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362987 A galvanometer of resistance $3663 Ω$ gives full scale deflection for a certain $I_{g}$ . Calculate the value of the resistance of the shunt which, when joined to the galvanometer coil will result in 1/ 34 of the total current passing through the galvanometer. Also the total resistance of the galvanometer and shunt will be

1

111 Ω, 107.7 Ω

2

110 Ω, 5 Ω

3

112 Ω, 105.7 Ω

4

112 Ω, 112 Ω

Explanation:

' $I$ ' is the total current then
$I_{g} = I / 34$
Now,
$S = \frac{I_{g} G}{I - I_{g}} = \frac{I_{g} \times 3663}{34 I_{g} - I_{g}} = \frac{3663 I_{g}}{33 I_{g}} = 111 Ω$
Combined resistance of shunt and galvanometer.
$R_{P} = \frac{G S}{G + S} = \frac{3663 \times 111}{3663 + 111} = \frac{406593}{3774} = 107.7 Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362988 A current of $200 μ A$ deflects the coil of a moving coil galvanometer through $60^{\circ} .$ The current to cause deflection through $\frac{π}{10}$ radian is :

1

180 μ A

2

120 μ A

3

30 μ A

4

60 μ A

Explanation:

Deflection in galvanometer is proportional to current flowing in the loop
As $I \propto θ$
$∴ \frac{I_{1}}{I_{2}} = \frac{θ_{1}}{θ_{2}}$
$\Rightarrow \frac{200 μ A}{I_{1}} = \frac{π / 3}{π / 10} = \frac{10}{3}$
$\Rightarrow I_{2} = 60 μ A$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362985 The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 %$ . The percentage change in voltage sensitivity of the galvanometer will be

1

0 %

2

50 %

3

75 %

4

100 %

Explanation:

Sensitivity of a moving coil galvanometer
$= \frac{N A B}{k}$
Deflection in current is increased but current sensitivity is independent of voltage sensitivity. Hence, percentage change in voltage sensitivity of the galvanometer is zero.
So, correct option is (1).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362986 For a moving coil galvanometer, the deflection in the coil is $0.05 r a d$ when a current of $10 m A$ is passed through it. If the torsional constant of suspension wire is $4.0 \times 10^{- 5} N m r a d^{- 1}$ , the magnetic field is $0.01 T$ and the number of turns in the coils is 200 , the area of each turn (in $c m^{2}$ ) is

1 2.0

2 1.0

3 1.5

4 0.5

Explanation:

$θ = 0.05 r a d$ , Current, $i = 10 m A$ ;
Torsional constant, $k = 4 \times 10^{- 5} N m / r a d$
$B = 0.01 T, N = 200$
Let area is $A$ .
At equilibrium
$τ = k θ$
$N i A B = k θ$
$A = \frac{k θ}{N i B} = \frac{4 \times 10^{- 5} \times 0.05}{200 \times 10 \times 10^{- 3} \times 0.01} m^{2}$
$A = 1 c m^{2}$
So, correction is (2).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362987 A galvanometer of resistance $3663 Ω$ gives full scale deflection for a certain $I_{g}$ . Calculate the value of the resistance of the shunt which, when joined to the galvanometer coil will result in 1/ 34 of the total current passing through the galvanometer. Also the total resistance of the galvanometer and shunt will be

1

111 Ω, 107.7 Ω

2

110 Ω, 5 Ω

3

112 Ω, 105.7 Ω

4

112 Ω, 112 Ω

Explanation:

' $I$ ' is the total current then
$I_{g} = I / 34$
Now,
$S = \frac{I_{g} G}{I - I_{g}} = \frac{I_{g} \times 3663}{34 I_{g} - I_{g}} = \frac{3663 I_{g}}{33 I_{g}} = 111 Ω$
Combined resistance of shunt and galvanometer.
$R_{P} = \frac{G S}{G + S} = \frac{3663 \times 111}{3663 + 111} = \frac{406593}{3774} = 107.7 Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362988 A current of $200 μ A$ deflects the coil of a moving coil galvanometer through $60^{\circ} .$ The current to cause deflection through $\frac{π}{10}$ radian is :

1

180 μ A

2

120 μ A

3

30 μ A

4

60 μ A

Explanation:

Deflection in galvanometer is proportional to current flowing in the loop
As $I \propto θ$
$∴ \frac{I_{1}}{I_{2}} = \frac{θ_{1}}{θ_{2}}$
$\Rightarrow \frac{200 μ A}{I_{1}} = \frac{π / 3}{π / 10} = \frac{10}{3}$
$\Rightarrow I_{2} = 60 μ A$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362985 The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 %$ . The percentage change in voltage sensitivity of the galvanometer will be

1

0 %

2

50 %

3

75 %

4

100 %

Explanation:

Sensitivity of a moving coil galvanometer
$= \frac{N A B}{k}$
Deflection in current is increased but current sensitivity is independent of voltage sensitivity. Hence, percentage change in voltage sensitivity of the galvanometer is zero.
So, correct option is (1).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362986 For a moving coil galvanometer, the deflection in the coil is $0.05 r a d$ when a current of $10 m A$ is passed through it. If the torsional constant of suspension wire is $4.0 \times 10^{- 5} N m r a d^{- 1}$ , the magnetic field is $0.01 T$ and the number of turns in the coils is 200 , the area of each turn (in $c m^{2}$ ) is

1 2.0

2 1.0

3 1.5

4 0.5

Explanation:

$θ = 0.05 r a d$ , Current, $i = 10 m A$ ;
Torsional constant, $k = 4 \times 10^{- 5} N m / r a d$
$B = 0.01 T, N = 200$
Let area is $A$ .
At equilibrium
$τ = k θ$
$N i A B = k θ$
$A = \frac{k θ}{N i B} = \frac{4 \times 10^{- 5} \times 0.05}{200 \times 10 \times 10^{- 3} \times 0.01} m^{2}$
$A = 1 c m^{2}$
So, correction is (2).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362987 A galvanometer of resistance $3663 Ω$ gives full scale deflection for a certain $I_{g}$ . Calculate the value of the resistance of the shunt which, when joined to the galvanometer coil will result in 1/ 34 of the total current passing through the galvanometer. Also the total resistance of the galvanometer and shunt will be

1

111 Ω, 107.7 Ω

2

110 Ω, 5 Ω

3

112 Ω, 105.7 Ω

4

112 Ω, 112 Ω

Explanation:

' $I$ ' is the total current then
$I_{g} = I / 34$
Now,
$S = \frac{I_{g} G}{I - I_{g}} = \frac{I_{g} \times 3663}{34 I_{g} - I_{g}} = \frac{3663 I_{g}}{33 I_{g}} = 111 Ω$
Combined resistance of shunt and galvanometer.
$R_{P} = \frac{G S}{G + S} = \frac{3663 \times 111}{3663 + 111} = \frac{406593}{3774} = 107.7 Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362988 A current of $200 μ A$ deflects the coil of a moving coil galvanometer through $60^{\circ} .$ The current to cause deflection through $\frac{π}{10}$ radian is :

1

180 μ A

2

120 μ A

3

30 μ A

4

60 μ A

Explanation:

Deflection in galvanometer is proportional to current flowing in the loop
As $I \propto θ$
$∴ \frac{I_{1}}{I_{2}} = \frac{θ_{1}}{θ_{2}}$
$\Rightarrow \frac{200 μ A}{I_{1}} = \frac{π / 3}{π / 10} = \frac{10}{3}$
$\Rightarrow I_{2} = 60 μ A$

JEE - 2024

PHXII04:MOVING CHARGES AND MAGNETISM

362985 The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 %$ . The percentage change in voltage sensitivity of the galvanometer will be

1

0 %

2

50 %

3

75 %

4

100 %

Explanation:

Sensitivity of a moving coil galvanometer
$= \frac{N A B}{k}$
Deflection in current is increased but current sensitivity is independent of voltage sensitivity. Hence, percentage change in voltage sensitivity of the galvanometer is zero.
So, correct option is (1).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362986 For a moving coil galvanometer, the deflection in the coil is $0.05 r a d$ when a current of $10 m A$ is passed through it. If the torsional constant of suspension wire is $4.0 \times 10^{- 5} N m r a d^{- 1}$ , the magnetic field is $0.01 T$ and the number of turns in the coils is 200 , the area of each turn (in $c m^{2}$ ) is

1 2.0

2 1.0

3 1.5

4 0.5

Explanation:

$θ = 0.05 r a d$ , Current, $i = 10 m A$ ;
Torsional constant, $k = 4 \times 10^{- 5} N m / r a d$
$B = 0.01 T, N = 200$
Let area is $A$ .
At equilibrium
$τ = k θ$
$N i A B = k θ$
$A = \frac{k θ}{N i B} = \frac{4 \times 10^{- 5} \times 0.05}{200 \times 10 \times 10^{- 3} \times 0.01} m^{2}$
$A = 1 c m^{2}$
So, correction is (2).

JEE - 2023

PHXII04:MOVING CHARGES AND MAGNETISM

362987 A galvanometer of resistance $3663 Ω$ gives full scale deflection for a certain $I_{g}$ . Calculate the value of the resistance of the shunt which, when joined to the galvanometer coil will result in 1/ 34 of the total current passing through the galvanometer. Also the total resistance of the galvanometer and shunt will be

1

111 Ω, 107.7 Ω

2

110 Ω, 5 Ω

3

112 Ω, 105.7 Ω

4

112 Ω, 112 Ω

Explanation:

' $I$ ' is the total current then
$I_{g} = I / 34$
Now,
$S = \frac{I_{g} G}{I - I_{g}} = \frac{I_{g} \times 3663}{34 I_{g} - I_{g}} = \frac{3663 I_{g}}{33 I_{g}} = 111 Ω$
Combined resistance of shunt and galvanometer.
$R_{P} = \frac{G S}{G + S} = \frac{3663 \times 111}{3663 + 111} = \frac{406593}{3774} = 107.7 Ω$

PHXII04:MOVING CHARGES AND MAGNETISM

362988 A current of $200 μ A$ deflects the coil of a moving coil galvanometer through $60^{\circ} .$ The current to cause deflection through $\frac{π}{10}$ radian is :

1

180 μ A

2

120 μ A

3

30 μ A

4

60 μ A

Explanation:

Deflection in galvanometer is proportional to current flowing in the loop
As $I \propto θ$
$∴ \frac{I_{1}}{I_{2}} = \frac{θ_{1}}{θ_{2}}$
$\Rightarrow \frac{200 μ A}{I_{1}} = \frac{π / 3}{π / 10} = \frac{10}{3}$
$\Rightarrow I_{2} = 60 μ A$

JEE - 2024

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