Acircular loop carrying a current of 2.0 a is in a magnetic field of 3.5 t. the loop has an area of 0.12 m2 and its plane is oriented at a 37° angle to the field. what is the magnitude of the magnetic torque on the loop?

Torque,  

Explanation:

It is given that,

Magnetic field, B = 3.5 T

Current, I = 2 A

Area of cross section, A = 0.12 m²

Angle between the plane and the magnetic field, θ = 37°

We need to find the magnitude of torque on the loop. It is given by :

Here, is the angle between the magnetic field and the normal to the coil.

So, the magnitude of the magnetic torque on the loop is 0.67 N-m. Hence, this is the required solution.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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