The voltage sensitivity of a galvanometer is given by: Voltage Sensitivity = Current Sensitivity/R If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is: 1.2/1.25 = 0.96 This indicates a 4% decrease in voltage sensitivity. Answer: (D) DecreRead more
The voltage sensitivity of a galvanometer is given by:
Voltage Sensitivity = Current Sensitivity/R
If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is:
1.2/1.25 = 0.96
This indicates a 4% decrease in voltage sensitivity.
Answer: (D) Decrease by 4%.
The magnetic field around a long, straight current-carrying wire is given by: B = μ0I/2πr Substituting values: B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is: (B) 1 × 10⁻⁵ T acting upwarRead more
The magnetic field around a long, straight current-carrying wire is given by:
B = μ0I/2πr
Substituting values:
B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T
Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is:
(B) 1 × 10⁻⁵ T acting upwards.
An α-particle projected along the Y-axis with velocity v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along theRead more
An α-particle projected along the Y-axis with velocity
v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along the X-axis, resulting in a helical trajectory with its axis parallel to the X-axis.
Answer: (C) Helical with its axis parallel to X-axis.
The magnetic field around a long, straight current-carrying wire is given by Ampere’s circuital law as: B = μ0I/2πr .Here: μ0 4π × 10⁻⁷Tm/A (permeability of free space) I =10 A (current) r = 0.10m (distance) Substituting values: B = 4π × 10⁻⁷× 10/2π × 0.10 = 2 × 10⁻⁵T To determine the direction ofRead more
The magnetic field around a long, straight current-carrying wire is given by Ampere’s circuital law as:
B = μ0I/2πr
.Here:
μ0 4π × 10⁻⁷Tm/A (permeability of free space)
I =10 A (current) r = 0.10m (distance)
Substituting values:
B = 4π × 10⁻⁷× 10/2π × 0.10 = 2 × 10⁻⁵T
To determine the direction of the magnetic field, use the right-hand thumb rule: Point the thumb in the direction of current (east to west) and curl the fingers around the wire. At a point due north, the magnetic field points downward. Hence, the correct answer is:
(A) 2 × 10⁻⁵ T, acting downwards.
The current sensitivity of a galvanometer increases by 20%. If its resistance also increases by 25%, the voltage sensitivity will
The voltage sensitivity of a galvanometer is given by: Voltage Sensitivity = Current Sensitivity/R If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is: 1.2/1.25 = 0.96 This indicates a 4% decrease in voltage sensitivity. Answer: (D) DecreRead more
The voltage sensitivity of a galvanometer is given by:
Voltage Sensitivity = Current Sensitivity/R
If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is:
1.2/1.25 = 0.96
This indicates a 4% decrease in voltage sensitivity.
Answer: (D) Decrease by 4%.
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A current of 5 A is flowing from east to west In a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due south on the table is :
The magnetic field around a long, straight current-carrying wire is given by: B = μ0I/2πr Substituting values: B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is: (B) 1 × 10⁻⁵ T acting upwarRead more
The magnetic field around a long, straight current-carrying wire is given by:
B = μ0I/2πr
Substituting values:
B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T
Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is:
(B) 1 × 10⁻⁵ T acting upwards.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
There are unifonn electric and magnetic fields in a region pointing along X-axis. An α-particle is projected along Y-axis with a velocity v. The shape of the trajectory will be :
An α-particle projected along the Y-axis with velocity v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along theRead more
An α-particle projected along the Y-axis with velocity
v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along the X-axis, resulting in a helical trajectory with its axis parallel to the X-axis.
Answer: (C) Helical with its axis parallel to X-axis.
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A current 10 A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due north on the table is
The magnetic field around a long, straight current-carrying wire is given by Ampere’s circuital law as: B = μ0I/2πr .Here: μ0 4π × 10⁻⁷Tm/A (permeability of free space) I =10 A (current) r = 0.10m (distance) Substituting values: B = 4π × 10⁻⁷× 10/2π × 0.10 = 2 × 10⁻⁵T To determine the direction ofRead more
The magnetic field around a long, straight current-carrying wire is given by Ampere’s circuital law as:
B = μ0I/2πr
.Here:
μ0 4π × 10⁻⁷Tm/A (permeability of free space)
I =10 A (current) r = 0.10m (distance)
Substituting values:
B = 4π × 10⁻⁷× 10/2π × 0.10 = 2 × 10⁻⁵T
To determine the direction of the magnetic field, use the right-hand thumb rule: Point the thumb in the direction of current (east to west) and curl the fingers around the wire. At a point due north, the magnetic field points downward. Hence, the correct answer is:
(A) 2 × 10⁻⁵ T, acting downwards.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/