A heavy conductor (mass m, length ?, resistance R ) is suspended by two springs

Question

A heavy conductor (mass m, length ?, 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 ?gure. A magnetic ?eld B~ is now imposed. The acceleration of gravity is 9.8 m/s^2 .

Which direction should the magnetic ?eld

point to have the most e?ect in raising or

lowering the heavy conductor?

1. Parallel to the conductor

2. The direction of the magnetic ?eld is

irrelevant.

3. Perpendicular to the page

4. Vertical and in the plane of the page

5. In the plane of the page at a 45 DEGREESangle tothe conductor

2.Assume a magnetic ?eld B~ points out of the

plane of the page (and thus perpendicular to

the plane of the circuit).

The direction of the magnetic force on the

heavy conductor is

1. down.

2. into the page.3. out of the page.

4. to the right.

5. to the left.

6. undetermined, since the conductor does

not feel a force.

7. up.

3.With the magnetic ?eld pointing in the samedirection as in the previous part, what happens to the magnetic force on the heavy conductor if the resistance R decreases toR/2?

1. force doubles

2. force remains zero

3. force quadruples

4. force halves

5. force is quartered

6. force stays the same

4.What direction should the magnetic ?eldpoint to be the most e?ective for lifting theheavy conductor?

1. out of the page

2. horizontal and to the right

3. into the page

4. horizontal and to the left

5. vertical and upward

6. vertical and downward

5.The acceleration of gravity is 9.8 m/s^2.What is the minimum magnetic ?eldstrength required to completely take theweight of the heavy conductor o? the springs ifa heavy conductor mass 1.4 kg, length 0.85 m,resistance 11 ? (ohms) is suspended by two springseach with spring constant 3.1 N/m, and connected to a battery with electric potential16 V as shown in the ?gure.

Answer in units of T

6.Now, consider the extension of the springsif the conductor is placed on them with nomagnetic ?eld present.Calculate the magnetic ?eld strength required to double the extension of the spring.

Answer in units of T

Explanation / Answer

Given the conductor is shown on a page. 1. Perpendicular to the page Fb+Fg = 0 iLXB = mg iLB = mg B = mg/iL B = mg/((V/R)*L) B = (1.8*9.8)/((17/12)*0.84)=14.824 T Note: Since the forces balance, spring energy has ideally turned into gravitational potential energy.

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