Let's carefully analyze and solve the problem step-by-step. --- ### **Given Data:** - **Generator rotation speed (n):** 401 rpm (revolutions per minute) - **Number of turns in coil (N):** 301 turns - **Dimensions of the coil:** 5.00 cm by 7.95 cm - **Desired emf (V_peak):** 23.6 V - **We need to find:** Magnetic field strength \( B \) --- ### **Step 1: Convert rotation speed to radians per second** The emf depends on the rate of change of magnetic flux, which is related to the coil’s rotation. \[ n = 401 \text{ rpm} \] Convert rpm to revolutions per second: \[ \text{Revolutions per second} = \frac{n}{60} = \frac{401}{60} \approx 6.683 \text{ rev/sec} \] Since \( 1 \text{ revolution} = 2\pi \text{ radians} \): \[ \omega = \text{angular velocity} = 2\pi \times \text{rev/sec} = 2\pi \times 6.683 \approx 2\pi \times 6.683 \] Calculate: \[ \omega \approx 2 \times 3.1416 \times 6.683 \approx 41.99 \text{ rad/sec} \] --- ### **Step 2: Find the area of the coil** Convert the dimensions to meters: \[ \text{length} = 5.00 \text{ cm} = 0.050 \text{ m} \] \[ \text{width} = 7.95 \text{ cm} = 0.0795 \text{ m} \] Area: \[ A = \text{length} \times \text{width} = 0.050 \times 0.0795 = 0.003975 \text{ m}^2 \] --- ### **Step 3: Use the emf formula for a rotating coil** The **peak emf** induced in a coil rotating in a magnetic field is given by: \[ V_{peak} = N \times B \times A \times \omega \] *This is derived from the emf induced in a rotating coil:* \[ V(t) = N B A \omega \sin(\omega t) \] where the **peak emf** is: \[ V_{peak} = N B A \omega \] --- ### **Step 4: Rearrange the formula to solve for \( B \)** \[ B = \frac{V_{peak}}{N \times A \times \omega} \] Plugging in the known values: \[ V_{peak} = 23.6 \text{ V} \] \[ N = 301 \] \[ A = 0.003975 \text{ m}^2 \] \[ \omega = 41.99 \text{ rad/sec} \] Calculate: \[ B = \frac{23.6}{301 \times 0.003975 \times 41.99} \] --- ### **Step 5: Compute the denominator** First, compute \( N \times A \): \[ 301 \times 0.003975 \approx 1.196 \] Next, multiply by \( \omega \): \[ 1.196 \times 41.99 \approx 50.23 \] --- ### **Step 6: Final calculation for \( B \)** \[ B = \frac{23.6}{50.23} \approx 0.470 \text{ Tesla} \] --- ### **### Final Answer:** \[ \boxed{ \text{Magnetic field strength } B \approx \boxed{0.470 \text{ Tesla}} } \] **This is the magnetic field strength needed to produce a 23.6 V peak emf at the given rotation speed.**