A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 Ais set up in the rod through the wires. (a) What magnetic field should be set up normal to the conductorin order that the tension in the wires is zero? (b) What will be the total tension in the wires if the direction ofcurrent is reversed keeping the magnetic field same as before?(Ignore the mass of the wires.) g = 9.8 m s–2.
Given data: The length of the rod is 0.45 m , the mass suspended by the wires is 60 g , the acceleration due to gravity is 9.8 m/s 2 and the current in the rod flowing through wire is 5 A .
a)
Since, the magnetic field is equal and opposite to the weight of the wire.
The magnetic field is given as,
B = m g I l
Where, the length of the rod is l , the mass suspended by the wires is m , the acceleration due to gravity is g , the current in the rod flowing through wire is I and the magnetic field is B .
By substituting the values in the above equation, we get
B = 60 × 10 − 3 × 9.8 5 × 0.45 B = 0.26 T
Thus, the magnetic field should be set up 0.26 T normal to the conductor and the direction of magnetic force is upward.
b)
Since, the direction of current is reversed.
The total tension in the wire is given as,
T = I B l + m g
Where, the total tension in the wire is T .
By substituting the given values in the above expression, we get
T = 0.26 × 5 × 0.45 + ( 60 × 10 − 3 ) × 9.8 = 1.176 N
Thus, the total tension in the wire is 1.176 N .
Given data: The length of the rod is
a)
Since, the magnetic field is equal and opposite to the weight of the wire.
The magnetic field is given as,
Where, the length of the rod is
By substituting the values in the above equation, we get
Thus, the magnetic field should be set up
b)
Since, the direction of current is reversed.
The total tension in the wire is given as,
Where, the total tension in the wire is
By substituting the given values in the above expression, we get
Thus, the total tension in the wire is
