Question

If the 0.573 m diameter current loop, still carrying a current of 2.26 A, experiences a torque of 0.250N⋅m when oriented at an angle of θ=23.1, what is the magnitude of the magnetic field, ∣ B ⃗ ∣ ∣ B ⃗ ∣?

Answer #1

Diameter of the loop d = 0.573

Hence the radius of the loop r = 0.573/2 = 0.2865 m

Now the current in the loop I = 2.26 A .

Torque experience by the loop T = 0.25 N-m

Which makes an angle of

Now the magnitude of magnetic field is B

N = number of turns of loop = 1

Now the A = area of the loop = π r^2

A = π * ( 0.2865)^2

A = 0.2578 m^2

Now by the formula

B = 1.093 T

This is the magnitude of magnetic field

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