A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of ( )/3. What is the x component of the torque on this loop when it is placed in a uniform magnetic field of ( )T?
A) 4.7 N ⋅ m
B) 3.1 N ⋅ m
C) 19 N ⋅ m
D) 9.4 N ⋅ m
E) 12 N ⋅ m

Question

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of (  )/3. What is the x component of the torque on this loop when it is placed in a uniform magnetic field of (  )T?
A) 4.7 N ⋅ m 
B) 3.1 N ⋅ m 
C) 19 N ⋅ m 
D) 9.4 N ⋅ m 
E) 12 N ⋅ m

Quizplus · Accepted Answer

The torque on a current loop in a magnetic field is given by the equation:
$tau = IABsin(theta)$
where I is the current, A is the area of the loop, B is the magnetic field, and theta is the angle between the area vector of the loop and the magnetic field vector.

We are given the radius of the loop (0.50 m), the current (3.0 A), and the magnetic field (1.5 T). We are also given the unit normal vector of the loop, which we can use to find the area vector of the loop:
$A = pi r^2 = pi (0.50 m)^2 = 0.7854 m^2$
$theta = cos^{-1}(n dot B/B) = cos^{-1}((11eb6adc_ee0c_6116_b0cf_7140429d4092_TB8245_11)/3/(1.5 T)) = 0.9553 rad$

Plugging these values into the torque equation, we get:
$tau = (3.0 A)(0.7854 m^2)(1.5 T)sin(0.9553 rad) \approx 9.4 Nm$

Therefore, the x component of the torque is 9.4 Nm, which corresponds to choice D.

A Circular Loop (Radius = 0