The acceleration of gravity is 9.8 m/s^2. What is the minimum magnetic field str

Question

The acceleration of gravity is 9.8 m/s^2. 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 1.7 kg, length 0.94 m, resistance 13 Ohm is suspended by two springs each with spring constant 4.2 N/m, and connected to a battery with electric potential 21 V as shown in the figure. Answer in units of T. Now, consider the extension of the springs if the conductor is placed on them with no magnetic field present. Calculate the magnetic field strength required to double the extension of the spring. Answer in units of T.

Explanation / Answer

Q13

Mass of conductor m = 1.7 kg

length of conductor L = 0.94 m

Resistance R = 13 ohm

Battery emf = 21 V

magnetic force acting on conductor ,upwards= i L B (current to the right and B in to the page)

gravitational force acting on conductor ,downwards = mg

mg = i L B

1.7 x 9.8 = (21 / 13 ) x0.94 x B

B = 10.97 T (into the page)

Q14

Initially B = 0

two springs are bearing the load of wire. So tension is mg / 2 = kx, where x is the change in length of each spring.

when B is switched on so that it is out of page (current is still to the right), Magneitc force on wire is iLB downwards

now mg / 2 + iLB / 2 = 2kx (extension is doubled)

iLB/2 = kx = mg /2

iLB = mg

B = 10.97 T (out of page)

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