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NCERT Class XII Chapter
Moving Charges and Magnetism
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Question : 21
Total: 28
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
Solution:  

Tension in the strings and magnetic force IBl balance the weight of wire.
2T + IBl = mg
(a) For tension in the wire to be zero.
IBl = mg
B =
mg
Il
 =
60×10−3×9.8
5×0.45
 = 0.26 T
(b) If the direction of current is now reversed, keeping the current and magnetic field same, then 2T = IBl + mg
2T = 2 mg = 2 × 60 × 10–3 × 9.8
Total tension = 1.176 N.
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