21. A heavy conductor (mass m, length l, resistance R) is suspended by two sprin

Question

21.

A heavy conductor (mass m, length l, resistance R) is suspended by two springs each with spring constant k, and connected to a battery with electric potential V as shown in the figure. A magnetic field is now imposed. The acceleration of gravity is 9.8 m/s2. The acceleration of gravity is 9.8 m/s . What is the minimum magnetic field strength required to completely take the weight of the heavy conductor off the springs if a heavy conductor mass 1.6 kg, length 0.65 m, resistance 12 Ohm is suspended by two springs each with spring constant 1.9 N/m, and connected to a battery with electric potential 20 V as shown in the figure. Answer in units of T

Explanation / Answer

When there is no magnetic field, then the extension of the spring is solely due to the weight of the rod.

Let 'xmg' be the extension in the spring just due to weight

2 k xmg = mg (As springs are in parallel)

xmg = mg/2k

Now, we want to remove the weight of the rod, so that there is no extension in the spring.

i.e. xnew = 0

Zero extension in both the springs will occur only if a zero force is acting on the rod i.e. the weight of the rod is balanced by the magnetic force

Magnetic Force needed = mg (upward)

mg = I B l (I = current in the wire)

mg = VBl/R (As resistence = R, I = V/R)

Brequired|min = mgR/Vl , pointing perpendicularly into the plane of paper

Brequired|min = (1.6 x 9.8 x 12)/(20 x 0.65)

= 14.474 T , pointing perpendicularly into the plane of paper

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