The figure shows the orientation of a rectangular loop consisting of 80 closely wrapped turns each carrying a current I. The magnetic field in the region is ( ) mT. The loop can turn about the y axis. If the direction between area and magnetic field is 30, a = 0.40 m, b = 0.30 m, and I = 8.0 A, what is the magnitude of the torque exerted on the loop? The answer is1.5Nm.
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To find the magnitude of the torque exerted on the loop, we can use the formula:
Torque = N * A * B * sinθ
where,
N = number of turns in the loop
A = area of the loop
B = magnetic field
θ = angle between the area vector of the loop and the magnetic field vector
From the given information, we have:
N = 80 turns
A = (a * b) = (0.40 m * 0.30 m) = 0.12 m^2 (area of a rectangular loop)
B = mT (unknown value)
θ = 30 degrees
Plugging the values into the formula, we have:
Torque = (80 * 0.12 m^2 * B * sin(30 degrees))
The magnitude of the torque exerted on the loop is given as 1.5 Nm. Therefore, we can set up the equation and solve for B:
1.5 Nm = (80 * 0.12 m^2 * B * sin(30 degrees))
Rearranging the equation, we have:
B = (1.5 Nm) / (80 * 0.12 m^2 * sin(30 degrees))
Evaluating the expression, we find:
B ≈ 1.04 mT
Therefore, the magnitude of the magnetic field in the region is approximately 1.04 mT.