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A flat, 179 179 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.41 cm 2 4.41 cm2 and the angle between its magnetic dipole moment and the field is 59.9 ∘ . 59.9∘. Find the strength B B of the magnetic field that causes a torque of 2.25 × 10 − 5 N ⋅ m 2.25×10−5 N⋅m to act on the loop when a current of 2.49 mA 2.49 mA flows in it.

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Answer:

The value of the magnetic field is  [tex]B =0.1423T[/tex]

Explanation:

From the question we are told that

              The number of turns is  [tex]N = 179[/tex]

               The area of the loop is [tex]A = 4.41cm^2 = \frac{4.41}{10000} = 0.000414m[/tex]

                 The angle is  [tex]\theta = 59^o[/tex]

               The torque  is  [tex]\tau =2.25 * 10^{- 5} N[/tex]

                The current is  [tex]I = 2.49\ mA[/tex]

The torque acting on the current carry loop is  mathematically represented as

                     [tex]\tau = B * I * N * A * sin \theta[/tex]

Where is the magnitude of the magnetic filed

Making B the subject

                     [tex]B= \frac{\tau}{I * N * A * sin\theta}[/tex]

Substituting values

                    [tex]B = \frac{2.25*10^{-5}}{2.49*10^{-3} * 179 * 0.000414 * sin (59)}[/tex]

                       [tex]=0.1423 T[/tex]

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