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The Magnetic Dipole Moment of a Coil Problem A rectangular coil of dimensions 5.40 cm ✕ 8.50 cm consists of 25 turns of wire and carries a current of 20.0 mA. A 0.350 T magnetic field is applied parallel to the plane of the loop. A. What is the magnitude of its magnetic dipole moment?B. What is the magnitude of the torque acting on the loop?

User Amir Kost
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1 Answer

7 votes

Answer:

a


\mu = 0.0023 \ A\cdot m^2

b


\tau = 0.00080 \ N \cdot m

Step-by-step explanation:

From the question we are told that

The dimensions of the rectangular coil is 5.40 cm ✕ 8.50 cm = 0.054 m X 0.085 m

The number of turns is
N = 25 \ turns

The current it is carrying is
I = 20 \ mA = 0.02 \ A

The magnetic field is
B = 0.350 \ T

Generally the magnitude of the magnetic dipole moment is mathematically represented as


\mu = N * I * A

Here A is the area which is mathematically represented as


A = 0.054 * 0.085

=>
A = 0.00459 \ m^2

So


\mu = 25 * 0.02 * 0.00459

=>
\mu = 0.0023 \ A\cdot m^2

Generally the magnitude of the torque acting on the loop is mathematically represented as


\tau = \mu * B

=>
\tau = 0.0023 * 0.350

=>
\tau = 0.00080 \ N \cdot m

User Brian Moeskau
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