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The magnetic moment of a short dipole is 1 A·m². What is the magnitude of the magnetic induction in air at 10 cm from the center of the dipole on a line making an angle of 30° from the axis of the dipole?

a. 1.0 T
b. 0.5 T
c. 0.866 T
d. 1.155 T

1 Answer

2 votes

Final answer:

The magnitude of the magnetic induction in air at 10 cm from the center of a dipole on a line at a 30° angle from the axis of the dipole is not given among the options, but it is calculated to be 0.005 T.

Step-by-step explanation:

To calculate the magnetic induction (also referred to as the magnetic field B) at 10 cm from the center of a short dipole, making an angle of 30° from the axis of the dipole, we can use the formula for the magnetic field due to a dipole at a point in space:

B = µ0/(4π) * (m×/(r³)),

where µ0 is the magnetic constant (permeability of free space), m is the magnetic moment, θ is the angle from the axis of the dipole, and r is the distance from the dipole. Plugging in the values (µ0 = 4π × 10−7 N·A−2, m = 1 A·m², θ = 30°, r = 0.1 m), we get:

B = (4π × 10−7)/(4π) * (1 × sin(30°))/(0.1³) = 10−7 * (1/2) / (10−6)

B = 5 × 10−3 T or 0.005 T

Therefore, none of the given options (a. 1.0 T, b. 0.5 T, c. 0.866 T, d. 1.155 T) match the correct magnitude of the magnetic induction.

User Kulin Choksi
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