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 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 m s–2.
Given:
Length of rod = 0.45 m
Mass of rod = 60 gm
Current = 5.0 A
Acceleration due to gravity, g = 9.8 m/s2
(a) Let magnetic field in conductor so that tension is zero = B
Tension will be zero if the force due to magnetic field and the weight is balanced.
We know that,
B × I × l = m × g …(1)
Where,
B = magnetic field
I = current through the conductor
l = length of conductor
m = mass of conductor
g = acceleration due to gravity
From equation (1) we have,
Now plugging the values of m, g, I and l in equation (1)
⇒ 
⇒ B = 0.26 T
So, a magnetic field of 0.26 Tesla has to be set up normal to the conductor in order to get zero tension in the supporting wire. Using Fleming’s left hand rule we find that the magnetic field applies a force in upward direction on the conductor which balances its weight.
(b) Let total tension in wire if the direction of current is reversed = P
According to question, the magnetic field is kept the same. Using Fleming’s left hand rule we find that Force due to magnetic field on the conductor acts downwards.
Writing the balancing equations:
P = B × I × l m × g …(2)
Where,
P = tension
B = magnetic field
I = current through the conductor
l = length of conductor
m = mass of conductor
g = acceleration due to gravity
Now, putting the values in equation (2) we get,
⇒ P = 0.26T × 5A × 0.45A 0.06Kg × 9.8ms-2
⇒ P = 1.17 N
The tension in wire is 1.17 N.
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