Question

A 0.270 m radius, 530–turn coil is rotated one–fourth of a revolution in 3.81 ms, originally having its plane perpendicular to a uniform magnetic field. Find the magnetic field strength needed to induce an average emf of 9400 V.

Answer 1

Answer 2

Calculation:

1. Given Data:

• Radius of coil, $r=0.270\text{m}$

• Number of turns, $N=530$

• Time for rotation, $\mathrm{\Delta }t=3.81\text{ms}=3.81\times 10^{-3}\text{s}$

• Average emf induced, $\mathcal{E}=9400\text{V}$

• The coil is rotated one-fourth of a revolution, so the change in angle is $\mathrm{\Delta }\theta =\frac{\pi }{2}$ radians (from $0$ to $90^{\circ }$).

• 
  

2. Calculate the Change in Magnetic Flux ($\mathrm{\Delta }\mathrm{\Phi }$):
   The flux through one turn of the coil changes from ${\mathrm{\Phi }}_{\text{initial}}=BA$ (when perpendicular to the field) to ${\mathrm{\Phi }}_{\text{final}}=0$ (when parallel to the field).

• Area of the coil, $A=\pi r^2=\pi (0.270{)}^2=0.229{\text{m}}^2$

• Change in flux per turn:

  $$\mathrm{\Delta }\mathrm{\Phi }=BA-0=BA$$

• Total change in flux for $N$ turns:

  $$\mathrm{\Delta }{\mathrm{\Phi }}_{\text{total}}=N\cdot BA$$

3. Relate emf to the Rate of Change of Flux:
   Faraday's Law states:

   $$\mathcal{E}=-\frac{\mathrm{\Delta }{\mathrm{\Phi }}_{\text{total}}}{\mathrm{\Delta }t}$$

   Ignoring the negative sign (as we are interested in magnitude):

   $$9400=\frac{NBA}{\mathrm{\Delta }t}$$

5. Solve for the Magnetic Field ($B$):

   $$B=\frac{\mathcal{E}\cdot \mathrm{\Delta }t}{N\cdot A}=\frac{9400\times 3.81\times 10^{-3}}{530\times 0.229}$$$$B=\frac{35.814}{121.37}\approx 0.078\text{T}$$

 Answer:
The required magnetic field strength is 0.078 Tesla