Answer:
The magnitude of the torque on the loop due to the magnetic field is [tex]4.7\times10^{-4}\ N-m[/tex].
Explanation:
Given that,
Diameter = 10 cm
Current = 0.20 A
Magnetic field = 0.30 T
Unit vector[tex]n=-0.60\hat{i}-0.080\hat{j}[/tex]
We need to calculate the torque on the loop
Using formula of torque
[tex]\tau=NIAB\sin\theta[/tex]
Where, N = number of turns
A = area
I = current
B = magnetic field
Put the value into the formula
[tex]\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}[/tex]
[tex]\tau=4.7\times10^{-4}\ N-m[/tex]
Hence, The magnitude of the torque on the loop due to the magnetic field is [tex]4.7\times10^{-4}\ N-m[/tex].
The magnitude of the torque on the loop due to the magnetic field is 1.571 x 10⁻³ Nm.
The torque on the loop is calculated by using the following formulas;
τ = NIBAsinθ
where;
A = πr²
A = πd²/4
A = π(0.1)²/4 = 7.855 x 10⁻³ m²
τ = 1 x 0.2 x (7.855 x 10⁻³) x sin90
τ = 1.571 x 10⁻³ Nm.
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