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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.
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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%
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A galvannometer having a resistance of 20 Ω and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is
The voltage range of a voltmeter is given by: V = (n × k) × (RG + R) Where: n = 30 divisions k = 0.005 A/division (figure of merit) Rg = 20 Ω (galvanometer resistance) V = 15 volts 15 = (30 × 0.005) × (20 + R) 15 = 0.15 × (20 + R) 20 + R = 15/0.15 = 100 R = 80Ω Answer: (C) 80 Ω For more visit here:Read more
The voltage range of a voltmeter is given by:
V = (n × k) × (RG + R)
Where:
n = 30 divisions
k = 0.005 A/division (figure of merit)
Rg = 20 Ω (galvanometer resistance)
V = 15 volts
15 = (30 × 0.005) × (20 + R)
15 = 0.15 × (20 + R)
20 + R = 15/0.15 = 100
R = 80Ω
Answer: (C) 80 Ω
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A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represents the electric fields and magnetic fields respectively, then this region of space may have:
A proton moving with constant velocity implies no net force acts on it. The force on a proton due to electric and magnetic fields is given by: F =q( E + v × B) For no change in velocity, 0F =0, which is possible if: E =0 and B =0 E =0, B ≠0 (if velocity is parallel to magnetic field) Answer: (A) E =Read more
A proton moving with constant velocity implies no net force acts on it. The force on a proton due to electric and magnetic fields is given by:
F =q( E + v × B)
For no change in velocity,
0F =0, which is possible if:
E =0 and B =0
E =0, B ≠0
(if velocity is parallel to magnetic field)
Answer: (A) E = 0, B = 0
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When a current of 5 mA is passed through galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the a be resistance to be put in series with the galvanometer to convert it into to voltmeter of range 0-10 V is
To convert a galvanometer to a voltmeter, the series resistance (R) is calculated as: V = I(Rg + R) Where: V = 10 volts I = 5 mA = 0.005 A Rg = 15Ω 10 = 0.005 × (15+R) 15 + R = 10/0.005 = 2000 R = 2000 −15 = 1985Ω Answer: (C) 1.985 × 10³ Ω For more visit here: https://www.tiwariacademy.com/ncert-solRead more
To convert a galvanometer to a voltmeter, the series resistance (R) is calculated as:
V = I(Rg + R)
Where:
V = 10 volts
I = 5 mA = 0.005 A
Rg = 15Ω
10 = 0.005 × (15+R)
15 + R = 10/0.005 = 2000
R = 2000 −15 = 1985Ω
Answer: (C) 1.985 × 10³ Ω
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