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The galvanometer can be converted into an ammeter by connecting an appropriate resistor of resistance R in series with it. At the full-scale deflection current(I) of 4 mA, the ammeter must measure a current of 6 A. The resistance of the galvanometer coil is G =    Since the resistor and galvanometer coil are connected in parallel the potential difference is the same across them. The...
The galvanometer can be converted into a voltmeter by connecting an appropriate resistor of resistance R in series with it. At the full-scale deflection current(I) of 3 mA the voltmeter must measure a Voltage of 18 V. The resistance of the galvanometer coil G =    The galvanometer can be converted into a voltmeter by connecting a resistor of resistance  in series with it.
The magnetic field inside the solenoid is given by  n is number of turns per unit length n=1500 m-1 Current in the wire Iw = 6 A Mass of the wire m = 2.5 g Length of the wire l = 2 cm The windings of the solenoid would support the weight of the wire when the force due to the magnetic field inside the solenoid balances  weight of the wire Therefore a current of 108.37 A in the solenoid...
The average force on each electron in the coil due to the magnetic field will be eVdB where Vd is the drift velocity of the electrons. The current is given by where n is the free electron density and A is the cross-sectional area. The average force on each electron is
As we know the torque on a current-carrying loop in a magnetic field is given by the following relation It is clear that the torque, in this case, will be 0 as the area vector is along the magnetic field only.
The magnetic field is   Current in the loop=12 A Area of the loop = lengthbreadth A=0.10.05 A=0.005 m2 Since the area vector is in the opposite direction of the magnetic field the torque on the loop is zero. The force on the loop is zero. The force on the loop in all the above cases is zero  as the magnetic field is uniform
The magnetic field is   Current in the loop=12 A Area of the loop = lengthbreadth A=0.10.05 A=0.005 m2 Since the area vector is along the direction of the magnetic field the torque on the loop is zero. The force on the loop is zero.
The magnetic field is   Current in the loop=12 A Area of the loop = lengthbreadth A=0.10.05 A=0.005 m2 The torque on the loop has a magnitude 0.018 Nm and at an angle of 240o from the positive-x direction. The force on the loop is zero.
The magnetic field is   Current in the loop=12 A Area of the loop = lengthbreadth A=0.10.05 A=0.005 m2 The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-x-direction. The force on the loop is zero.
The magnetic field is   Current in the loop=12 A Area of the loop = lengthbreadth A=0.10.05 A=0.005 m2  (same as that in the last case) The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-y-direction. The force on the loop is zero. This was exactly the case in 24. (a) as well.
The magnetic field is   Current in the loop=12 A Area of the loop = lengthbreadth A=0.10.05 A=0.005 m2 The torque on the loop has a magnitude of 0.018 Nm and acts along the negative y-direction. The force on the loop is zero.
Magnetic field strength =1.5 T. Current flowing through the wire=7.0 A The angle between the direction of the current and magnetic field=45o The radius of the cylindrical region=10.0 cm The length of wire inside the magnetic field,   Force on a wire in a magnetic field is calculated by relation, F=2.1 N This force due to the magnetic field inside the cylindrical region acts on the wire in the...
The length of wire inside the magnetic field is equal to the diameter of the cylindrical region=20.0 cm=0.2 m. Magnetic field strenth=1.5 T. Current flowing through the wire=7.0 A The angle between the direction of the current and magnetic field=90o Force on a wire in a magnetic field is calculated by relation, This force due to the magnetic field inside the cylindrical region acts on the...
Since the distance between the wires is much smaller than the length of the wires we can calculate the force per unit length on the wires using the following relation. Current in both wires=300 A Distance between the wires=1.5 cm Permeability of free space=410-7 TmA-1 F=1.2 Nm-1
If the direction of the current is reversed the magnetic force would act in the same direction as that of gravity. Total tension in wires(T)=Gravitational force on rod + Magnetic force on the rod The total tension in the wires will be 1.176 N.
In order for the tension in the wires to be zero the force due to the magnetic field must be equal to the gravitational force on the rod. mass of rod=0.06 g length of rod=0.45m the current flowing through the rod=5 A A magnetic field of strength 0.261 T should be set up normal to the conductor in order that the tension in the wires is zero
The electron has been accelerated through a potential difference of 2.0 kV.  Therefore K.E of electron = 1.610-192000=3.210-16 J The component of velocity perpendicular to the magnetic field is The electron will move in a helical path of radius r given by the relation, r=5m r=510-4 m r=0.5 mm The component of velocity along the magnetic field is The electron will move in a helical path of...
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