A heavy conductor (mass m, length `, resis- tance R) is suspended by two springs

Question

A heavy conductor (mass m, length `, resis- tance 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 ~B is now imposed. The acceleration of gravity is 9.8 m/s2 .Which direction should the magnetic field point to have the most effect in raising or lowering the heavy conductor? 1. Parallel to the conductor 2. The direction of the magnetic field 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 angle to the conductor

Explanation / Answer

The force downward is mg. The forces upward and opposite mg are 2kx (that is: F(net spring) = k1*x + k2*x, and k1 and k2 are equal), and I*L x B. This is my reasoning: in order for the weight to be completely taken off of the springs, the upward force F(b) must equal the downward force, which is mg. Therefore, the spring force does not need to be taken into consideration. mg = I*L x B mg = (I * L-sub-x * i-hat) x (B-sub-z * k-hat)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---