(A) is correct. In a cyclotron, a charged particle undergoes continuous acceleration. It speeds up between the dees due to an alternating electric field but moves in circular motion inside the dees at constant speed due to the magnetic field. The repeated acceleration increases its energy, making itRead more
(A) is correct. In a cyclotron, a charged particle undergoes continuous acceleration. It speeds up between the dees due to an alternating electric field but moves in circular motion inside the dees at constant speed due to the magnetic field. The repeated acceleration increases its energy, making it spiral outward.
(A) is correct. The time period T of a charged particle in a uniform magnetic field is given by T = 2πm/qB . It depends on the mass m, charge q and magnetic field B, but not on the speed of the particle. Thus, the time period remains constant regardless of speed changes. For more visit here: https:/Read more
(A) is correct. The time period T of a charged particle in a uniform magnetic field is given by
T = 2πm/qB . It depends on the mass m, charge q and magnetic field B, but not on the speed of the particle. Thus, the time period remains constant regardless of speed changes.
(A) is correct. The force per unit length between two parallel current-carrying wires is given by F ∝ I₁ I₂/d, where d is the distance between them. If the distance is doubled, the force becomes 1/2 of the original value. Thus, the force between the wires is halved. For more visit here: https://www.Read more
(A) is correct. The force per unit length between two parallel current-carrying wires is given by
F ∝ I₁ I₂/d, where d is the distance between them. If the distance is doubled, the force becomes
1/2 of the original value. Thus, the force between the wires is halved.
The correct answer is (B) B. The magnetic field inside a long solenoid is given by: B = μonI Where n = number of turns per unit length and I = current. If the number of turns is halved (n/2) and current is doubled (2I), the new magnetic field becomes: B' = μ0 n/2 2I = μ0 nI = B Answer: (B) B. httRead more
The correct answer is (B) B.
The magnetic field inside a long solenoid is given by:
B = μonI
Where n = number of turns per unit length and
I = current.
If the number of turns is halved (n/2) and current is doubled (2I), the new magnetic field becomes:
B’ = μ0
n/2 2I = μ0
nI = B
Answer: (B) B.
The current sensitivity of a moving coil galvanometer is given by: Si = N AB/K Where: N = Number of turns A = Area of coil B = Magnetic field K= Torsional constant When the resistance (R) of the galvanometer increases by a factor of 3, the voltage sensitivity (Sv) is given by: Sv = Si/R Since currenRead more
The current sensitivity of a moving coil galvanometer is given by:
Si = N AB/K
Where:
N = Number of turns
A = Area of coil
B = Magnetic field
K= Torsional constant
When the resistance (R) of the galvanometer increases by a factor of 3, the voltage sensitivity (Sv) is given by:
Sv = Si/R
Since current sensitivity increases by 35%, we have:
Sv′ = 1.35 × Si/3R = 1.35/3 × Sv = 0.45 × Sv
Hence, the voltage sensitivity decreases by 55% (100%−45%
Answer: (C) 55%
In acyclotron, a charged particle
(A) is correct. In a cyclotron, a charged particle undergoes continuous acceleration. It speeds up between the dees due to an alternating electric field but moves in circular motion inside the dees at constant speed due to the magnetic field. The repeated acceleration increases its energy, making itRead more
(A) is correct. In a cyclotron, a charged particle undergoes continuous acceleration. It speeds up between the dees due to an alternating electric field but moves in circular motion inside the dees at constant speed due to the magnetic field. The repeated acceleration increases its energy, making it spiral outward.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
Time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of
(A) is correct. The time period T of a charged particle in a uniform magnetic field is given by T = 2πm/qB . It depends on the mass m, charge q and magnetic field B, but not on the speed of the particle. Thus, the time period remains constant regardless of speed changes. For more visit here: https:/Read more
(A) is correct. The time period T of a charged particle in a uniform magnetic field is given by
T = 2πm/qB . It depends on the mass m, charge q and magnetic field B, but not on the speed of the particle. Thus, the time period remains constant regardless of speed changes.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
If distance between two curreni-carrying wires is doubled, then force between them is
(A) is correct. The force per unit length between two parallel current-carrying wires is given by F ∝ I₁ I₂/d, where d is the distance between them. If the distance is doubled, the force becomes 1/2 of the original value. Thus, the force between the wires is halved. For more visit here: https://www.Read more
(A) is correct. The force per unit length between two parallel current-carrying wires is given by
F ∝ I₁ I₂/d, where d is the distance between them. If the distance is doubled, the force becomes
1/2 of the original value. Thus, the force between the wires is halved.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
A long solenoid carrying current produces a magnetic field B along its axis. If the number of turns in the solenoid is halved and current in it is doubled the new magnetic field will be :
The correct answer is (B) B. The magnetic field inside a long solenoid is given by: B = μonI Where n = number of turns per unit length and I = current. If the number of turns is halved (n/2) and current is doubled (2I), the new magnetic field becomes: B' = μ0 n/2 2I = μ0 nI = B Answer: (B) B. httRead more
The correct answer is (B) B.
The magnetic field inside a long solenoid is given by:
B = μonI
Where n = number of turns per unit length and
I = current.
If the number of turns is halved (n/2) and current is doubled (2I), the new magnetic field becomes:
B’ = μ0
n/2 2I = μ0
nI = B
Answer: (B) B.
https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
See lessThe current sensitivity of a moving coil galvanometer increases by 35%, when its resistance is increased by a factor 3. The voltage sensitivity of galvanometer changes by a factor
The current sensitivity of a moving coil galvanometer is given by: Si = N AB/K Where: N = Number of turns A = Area of coil B = Magnetic field K= Torsional constant When the resistance (R) of the galvanometer increases by a factor of 3, the voltage sensitivity (Sv) is given by: Sv = Si/R Since currenRead more
The current sensitivity of a moving coil galvanometer is given by:
Si = N AB/K
Where:
N = Number of turns
A = Area of coil
B = Magnetic field
K= Torsional constant
When the resistance (R) of the galvanometer increases by a factor of 3, the voltage sensitivity (Sv) is given by:
Sv = Si/R
Since current sensitivity increases by 35%, we have:
Sv′ = 1.35 × Si/3R = 1.35/3 × Sv = 0.45 × Sv
Hence, the voltage sensitivity decreases by 55% (100%−45%
Answer: (C) 55%
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/