A uniform rectangular coil of total mass 212 g and dimensions 0.500 m ×1.00 m is oriented wit

A uniform rectangular coil of total mass 212 g and...

A uniform rectangular coil of total mass 212$\mathrm { g }$ and dimensions 0.500$\mathrm { m } \times$ 1.00$\mathrm { m }$ is oriented with its plane parallel to a uniform $3.00 - \mathrm { T }$ magnetic field (Fig. E27.47). A current of 2.00$\mathrm { A }$ is suddenly started in the coil. (a) About which axis $( A _ { 1 }$ or $A _ { 2 })$ will the coil begin to rotate? Why? (b) Find the initial angular acceleration of the coil just after the current is started.

A uniform rectangular coil of total mass 212$\mathrm { g }$ and dimensions 0.500$\mathrm { m } \times$
1.00$\mathrm { m }$ is oriented with its plane parallel to a uniform $3.00 - \mathrm { T }$
magnetic field (Fig. E27.47). A current of 2.00$\mathrm { A }$ is suddenly started in the coil. (a) About which axis $( A _ { 1 }$ or $A _ { 2 })$ will the coil begin to rotate? Why?
(b) Find the initial angular acceleration of the coil just after the current is started.

Key Concepts

Rotational Dynamics and Moment of Inertia

Once a torque is applied, the rotational response of an object is determined by its moment of inertia, which encapsulates how the mass is distributed with respect to the rotation axis. Newton’s second law for rotation, expressed as ? = I? (with ? being the angular acceleration), connects the torque produced by the magnetic forces to the resulting rotational motion of the loop. Computing the moment of inertia correctly for the specific geometry and axis is critical for predicting the angular acceleration.

Magnetic Torque on Current Loops

When a current-carrying loop is placed in a magnetic field, the field exerts forces on the different segments of the loop. Although the net force on a closed loop in a uniform magnetic field is generally zero, these forces can produce a torque that tends to rotate the loop so that its magnetic moment aligns with the magnetic field. The expression ? = ? × B, where ? (the magnetic moment) is proportional to the product of the current and the loop area, is key to understanding this behavior.

Lorentz Force on Current-Carrying Conductors

The Lorentz force law, F = I(L × B), describes the force experienced by a segment of conductor carrying a current in a magnetic field. This force is responsible for the individual contributions that, when applied to different parts of the loop, result in a net torque. Understanding this force is essential for analyzing how the magnetic field interacts with the current distribution in the loop, leading to rotational effects.

Transcript

00:01 The magnetic field exerts a torque on the current carrying coil which causes us to turn.

00:08 And we can find whether the coil will rotate about axis a1 or a2 using this formula.

00:19 So basically torque is equal to mu cross a .b where mu is given by the magnitude of m.

00:33 Is given by i times a the current so for this particular case you should point if the current is clockwise then you should point out of the paper and if current is anti -clockwise you should point the direction of torque ultimately will be so there will be no talk about a1 because yeah there won't be any because the cross product will just be zero but there will be a talk about a2 because only the torque is along top and bottom sides of the coil so find the moment to find the angular acceleration of the coil we use newton's second law which says the total torque is equal to i times the angular acceleration so now we know the to find to find now we know that first we have to find the moment of inertia so to find that treat the two one meter segments this and this as point masses since all the points in them are 0 .25 meter from the rotation axis a 2 now the 2 0 .5 segments 5 meter segments are treat them as thin uniform bars rotated about their centers so so this segment and this segment are basically 2 rods rotating about axis a2, so rotating about their center.

02:43 And a, b, and b and d .c are rods, but they can be treated as point masses, which are rotating about the axis a2, that is at a distance of 0 .25 meters.

02:59 So this is this distance.

03:01 So that's half of this distance.

03:03 So 2 .25 meters.

03:07 Now, we know that since the coil is uniform, the mass of each segment is proportional to its fraction of the total perimeter of the coil.

03:19 So each one meter segment is one third of the total perimeter.

03:25 So its mass will be.

03:31 So let's find the total perimeter first.

03:34 So total perimeter is twice of this plus this.

03:39 So that's basically three meters.

03:42 So each one meter segment will have a mass then the total mass.

03:57 So basically one third of the total mass.

04:01 So one comes from the length of the segment, three is the total perimeter...