A single-turn current loop, carrying a current of 4.00 A, is in the shape of a right triangle wi

A single-turn current loop, carrying a current of 4.00 A, is...

A single-turn current loop, carrying a current of $4.00 \mathrm{A},$ is in the shape of a right triangle with sides $50.0,120,$ and 130 $\mathrm{cm} .$ The loop is in a uniform magnetic field of magnitude 75.0 $\mathrm{mT}$ whose direction is parallel to the current in the 130 $\mathrm{cm}$ side of the loop. What is the magnitude of the magnetic force on (a) the 130 $\mathrm{cm}$ side, (b) the 50.0 $\mathrm{cm}$ side, and $(\mathrm{c})$ the 120 $\mathrm{cm}$ side? (d) What is the magnitude of the net force on the loop?

A single-turn current loop, carrying a current of $4.00 \mathrm{A},$ is in the
shape of a right triangle with sides $50.0,120,$ and 130 $\mathrm{cm} .$ The loop
is in a uniform magnetic field of magnitude 75.0 $\mathrm{mT}$ whose direction is parallel to the current in the 130 $\mathrm{cm}$ side of the loop. What is
the magnitude of the magnetic force on (a) the 130 $\mathrm{cm}$ side, (b) the
50.0 $\mathrm{cm}$ side, and $(\mathrm{c})$ the 120 $\mathrm{cm}$ side? (d) What is the magnitude of the net force on the loop?

Key Concepts

Magnetic Force on a Current-Carrying Conductor

This concept involves calculating the force experienced by a conductor carrying an electric current when it is placed in a magnetic field. The force is given by the equation F = I (L × B), where I is the current, L is the vector representing the length and direction of the conductor, and B is the magnetic field vector. This formula is central to understanding how magnetic fields interact with currents, determining the magnitude and direction of the force on individual wire segments.

Cross Product in Electromagnetism

The cross product is a vector operation used in the Lorentz force equation to determine the direction and magnitude of the force on a current element. It inherently incorporates the right-hand rule, which is essential for predicting the orientation of the resulting force relative to the directions of both the current and the magnetic field. This mathematical operation ensures that only the component of the current-carrying segment that is perpendicular to the magnetic field contributes to the force.

Uniform Magnetic Field

A uniform magnetic field is one in which the magnetic field strength and direction remain constant throughout the region of interest. In the context of current loops, a uniform magnetic field allows for simplified calculations of magnetic forces on each segment, since the field does not vary along the path of the current. This uniformity also leads to interesting results, such as the cancellation of net forces on closed current loops while still possibly producing a net torque.

Net Force on a Closed Current Loop

When a complete closed loop carrying a current is placed in a uniform magnetic field, the individual forces on each segment, while non-zero, typically sum to zero due to their vector nature. This cancellation occurs because the forces on opposite sides of the loop are equal in magnitude and opposite in direction, leading to a zero net translational force on the loop, even though a net torque can be present depending on the loop's orientation.

Transcript

00:01 In this problem, we're told to consider a single -turn loop carrying a current of 4 amps.

00:04 It's in the shape of a right triangle, and i've written a diagram down here with all the pertinent information.

00:10 The loop is in a uniform magnetic field of magnitude 75 .0 mt, and the direction is parallel to the current in the 130 -centimeter side of the loop.

00:22 So for a, we're asked to find the magnitude of the force on the 130 centimeter side.

00:36 I'll make sure i'm doing the right one.

00:47 That'll be given by my force is i times l times b times my sine of theta.

01:02 So this will be my force will be equal to i times c times b times my sine of 90 degrees and we can see that this force will be equal to zero.

01:31 Next we have the magnitude on the 50 centimeter side.

01:36 So in this the tangent theta will be b over a so theta will be tangent of b over a that'll equal tangent will be 120 centimeters over 50 centimeters that'll equal 67 .4 degrees okay now we can do our fourth same equation as our last page of a will equal 4 .00 amps times, next i'll have 50 .0 times 10 to the minus 2 meters times 75 .0 times 10 to the minus 3 t's times the sine of 67 .4 degrees.

02:43 And this will equal 0 .138 newtons.

02:52 And that will be our answer for b.

02:56 Let's go to c.

03:00 C, the magnitude of the force on the the 120 centimeter side.

03:10 This will be similar...

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