A rigid circular loop having a radius of 0.20 m and is in the xy-plane, carries a uniform external magnetic field. B = 0.20 T and is directed parallel to the plane of the loop. The torque exerted on the loop is given by the formula as follows;τ = NABsin θ
Where τ is the torque, N is the number of turns, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the normal to the loop and the magnetic field direction. The direction of the torque is given by the right-hand rule and depends on the direction of the magnetic field.To obtain the magnitude of the torque, we need to evaluate all the terms in the formula, τ = NABsinθ.A = πr² = π (0.20 m)² = 0.1257 m²The angle between the normal to the loop and the magnetic field direction is 0 degrees since the field is parallel to the loop.θ = 0°Substituting the values in the formula, we have;τ = NABsinθ= N (0.1257 m²) (0.20 T) (sin 0°)τ = 0 NmThe torque exerted on the loop is zero since the angle between the normal to the loop and the magnetic field direction is zero. The torque only exists when there is an angle between the normal and the magnetic field direction. Therefore, there is no net force on the current loop.For such more question on torque
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A new ride being built at an amusement park includes a vertical drop of 79.8 meters. Starting from rest, the ride vertically drops that distance before the track curves forward. If friction is neglected, what would be the speed of the roller coaster at the bottom of the drop?
Answer:
Approximately [tex]39.6\; {\rm m\cdot s^{-1}}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Under the assumptions, vertical acceleration of the vehicle during the ride would be equal to the gravitational field strength: [tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Apply the following SUVAT equation to find the velocity of the vehicle at the bottom of the drop:
[tex]v^{2} - u^{2} = 2\, a\, x[/tex],
Where:
[tex]v[/tex] is the final velocity at the bottom of the drop;[tex]u[/tex] is the initial velocity at the beginning of the drop; [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] since the vehicle started from rest;[tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex] is the vertical acceleration of the vehicle during the drop;[tex]x = 79.8\; {\rm m}[/tex] is the vertical displacement of the vehicle during the drop.Rearrange this equation to find [tex]v[/tex]:
[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x} \\ &\approx \sqrt{0^{2} + 2\, (9.81)\, (79.8)} \; {\rm m\cdot s^{-1}} \\ &\approx 39.6\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the speed of this vehicle at the bottom of the drop would be approximately [tex]39.6\; {\rm m\cdot s^{-1}}[/tex].
17. The time taken by a 100 g stone to reach the ground when dropped from the top of (A) your school building has been given. Write a suitable formula for the calculation of ght of your school building.
