To find the angle between the magnetic field and the normal to the coil, we can use the formula for magnetic flux:
\[ \Phi = B A \cos(\theta) \]
where:
- \(\Phi\) is the magnetic flux (in webers, Wb),
- \(B\) is the magnetic field strength (in teslas, T),
- \(A\) is the area of the coil (in square meters, m²),
- \(\theta\) is the angle between the magnetic field and the normal to the coil (in degrees or radians).
Step 1: Calculate the Area of the Coil
The area \(A\) of the circular coil can be calculated using the formula:
\[ A = \pi r^2 \]
Given that the radius \(r = 1.5 , m\):
\[ A = \pi (1.5)^2 = \pi \cdot 2.25 \approx 7.06858 , m^2 \]
Step 2: Rearranging the Flux Equation
We can rearrange the magnetic flux formula to solve for \(\cos(\theta)\):
\[ \cos(\theta) = \frac{\Phi}{B \cdot A} \]
Step 3: Substitute the Known Values
Now, substitute \(\Phi = 6.04 , Wb\), \(B = 3.3 , T\), and \(A \approx 7.06858 , m^2\):
\[ \cos(\theta) = \frac{6.04}{3.3 \cdot 7.06858} \]
Calculating the denominator:
\[ 3.3 \cdot 7.06858 \approx 23.327 \]
Now substitute this value into the equation:
\[ \cos(\theta) = \frac{6.04}{23.327} \approx 0.259 \]
Step 4: Finding the Angle
Now we can use the inverse cosine function to find \(\theta\):
\[ \theta = \cos^{-1}(0.259) \]
Calculating this gives:
\[ \theta \approx 75.3^\circ \]
Final Answer
Rounding to two decimal places, the angle \(\theta\) between the magnetic field and the normal to the coil is:
\[ \theta \approx 75.30^\circ \]
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