Question

In the arrangement shown in the figure, the restore is 7 Omega and a 9 T magnetic field is directed out of the paper. The separation between the rails is 6 m. An applied force moves the bar to the left at a constant speed of 6 m/s.a. Calculate the applied force required to move the bar to the left at a constant speed of 6 m/s. assume the bar and rails have negligible resistance and friction. Neglect the mass of the bar. Answer in units of N.

b. At what rate is energy dissipated in the resistor? Answer in unts of W.

Answer 1

Final Answers:

The applied force required to move the bar to the left at a constant speed of 6 m/s is 126 N

Step-by-step explanation:

Calculating the applied force: The formula to find the force in this scenario involves the relationship between magnetic force and the current passing through the conductor. As the bar moves at a constant speed without acceleration, the magnetic force is balanced by the applied force. Using the formula F = BIL, where F is the magnetic force, B is the magnetic field, I is the current, and L is the length of the conductor, rearranging the equation for I gives I = F / (BL). The resistance R = 7 Ω, and according to Ohm's law, V = IR. Hence, the voltage V = IR = 7 Ω * I. Also, V = B * L * v, where v is the speed of the bar. Equating these voltage equations gives 7 Ω * I = B * L * v, and solving for I yields I = (B * L * v) / 7 Ω. Given B = 9 T, L = 6 m, and v = 6 m/s, substituting these values into the equation gives I = (9 T * 6 m * 6 m/s) / 7 Ω, resulting in I = 54 A. Then, using the rearranged formula I = F / (BL), the applied force F = I * B * L = 54 A * 9 T * 6 m = 126 N.

Final answer:

The rate at which energy is dissipated in the resistor is 756 W.

Step-by-step explanation:

Determining the energy dissipation rate:** The power dissipated in a resistor is given by the formula P = I^2 * R, where P is power, I is current, and R is resistance. As calculated previously, the current passing through the resistor is 54 A, and the resistance is 7 Ω. Substituting these values into the formula gives P = (54 A)^2 * 7 Ω = 756 W. Therefore, the rate at which energy is dissipated in the resistor is 756 watts. This power represents the rate of energy conversion from electrical energy to heat energy within the resistor due to the passage of current through it.

Answer 2

a. The applied force required to move the bar to the left at a constant speed of 6 m/s is 324 N. b. The rate at which energy is dissipated in the resistor is 252 W.

Step-by-step explanation:

a. To calculate the applied force required to move the bar to the left at a constant speed of 6 m/s, we need to consider the magnetic force acting on the bar. The magnetic force can be calculated using the equation F = BIL, where F is the force, B is the magnetic field, I is the current, and L is the length of the conductor in the magnetic field. Since the bar is moving at a constant speed and there is no resistance or friction, the applied force is equal to the magnetic force. Rearranging the equation to solve for I, we have I = F / (BL). Plugging in the given values, we find I = (F) / (9 T * 6 m). As the bar has negligible resistance, the induced current flows through the circuit, resulting in a magnetic force that opposes the applied force. Therefore, we can rearrange the equation to solve for F, which gives us F = I * (B * L). Plugging in the given values, we find F = (6 m/s) * (9 T * 6 m) = 324 N.

b. To calculate the rate at which energy is dissipated in the resistor, we need to calculate the power dissipated. Power is calculated using the equation P = I^2 * R, where P is the power, I is the current, and R is the resistance. By using the previously calculated value for the current, we can calculate the power. Plugging in the given values, we have P = (6 m/s)^2 * 7 Ω = 252 W.