A 25.0 -cm-long metal rod lies in the x y -plane and makes...

A 25.0 -cm-long metal rod lies in the $x y$ -plane and makes an angle of $36.9^{\circ}$ with the positive $x$ -axis and an angle of $53.1^{\circ}$ with the positive $y$ -axis. The rod is moving in the $+x$ -direction with a speed of 6.80 $\mathrm{m} / \mathrm{s}$ . The rod is in a uniform magnetic field $\vec{B}=(0.120 \mathrm{T}) \hat{i}-(0.220 \mathrm{T}) \hat{J}-(0.0900 \mathrm{T}) \hat{k}$ (a) What is the magnitude of the emf induced in the rod? (b) Indicate in a sketch which end of the rod is at higher potential.

A 25.0 -cm-long metal rod lies in the $x y$ -plane and makes an angle of $36.9^{\circ}$ with the positive $x$ -axis and an angle of $53.1^{\circ}$ with the positive $y$ -axis. The rod is moving in the $+x$ -direction with a speed of 6.80 $\mathrm{m} / \mathrm{s}$ . The rod is in a uniform magnetic field
$\vec{B}=(0.120 \mathrm{T}) \hat{i}-(0.220 \mathrm{T}) \hat{J}-(0.0900 \mathrm{T}) \hat{k}$ (a) What is the magnitude of the emf induced in the rod? (b) Indicate in a sketch which end of the rod is at higher potential.

Key Concepts

Motional EMF

Motional electromotive force (emf) refers to the voltage generated across a conductor when it moves through a magnetic field. This phenomenon occurs because the movement causes free charges in the conductor to experience a magnetic force, leading to a separation of charge and subsequently an induced voltage. It is a fundamental concept in electromagnetic induction and lies at the heart of many practical applications such as electric generators.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving in both electric and magnetic fields. In the context of a moving conductor, it is the magnetic component of the Lorentz force that acts on the free charges (F = q(v × B)). This force causes the charge separation that gives rise to the motional emf. Understanding the Lorentz force is crucial for predicting how the movement of conductors through magnetic fields will give rise to electrical effects.

Vector Cross Product

The vector cross product is a mathematical operation used to determine the magnitude and direction of the resulting vector when two vectors are multiplied. In problems involving motional emf, the cross product of the velocity vector and the magnetic field vector (v × B) determines the direction and magnitude of the force acting on a charge. This operation is essential for correctly applying the formula for motional emf and for analyzing the geometric relationships between the vectors involved.

Right-Hand Rule

The right-hand rule is a mnemonic used in physics to determine the direction of the vector resulting from a cross product. In electromagnetic induction, it helps in identifying the direction of the induced current or the orientation of the induced potential difference by aligning the fingers in the direction of the first vector (velocity) and curling them toward the second vector (magnetic field). The thumb then points in the direction of the resulting vector, such as the force or emf, making it an essential tool in visualizing and solving problems in electromagnetism.

Transcript

00:02 Hi, in the given problem the metallic rod is oriented in xy plane as shown in the figure.

00:15 Here this is the rod oriented in xy plane such that it is making an angle of 36 .9 degree with the x -axis and the length of the rod is given as 25 .0 centimeter or we can say this is 0 .25 meter so in vector form this length can be given as 0 .25 meter having a component along x -axis as cosine of 36 .9 degree i cap and a component along y -axis as sign 36 .9 degree j cap.

01:10 So when we rearrange these terms we get the length of this rod in vector form as it comes out to be this is 0 .2 i cap plus 0 .15 jcap so this is the length of the rod in vector form.

01:41 Now the rod is moving along x -axis with the velocity v, which is given as it is having a magnitude of 6 .80 meter per second and as it is moving along x -axis, so in vector form it can be given as 6 .80 meter per second i gap.

02:05 Now, first of all, we have to find a value of the emf induced across this rod, the magnitude of the emf induced.

02:19 So the expression for this emf induced is given as it is found by cross product of velocity vector with the magnetic field vector, which is then having a dot product with the length vector.

02:35 So first of all we will find the cross product of velocity vector with magnetic field vector.

02:45 So we will put these values.

02:46 For the velocity this is 6 .80 icap cross.

02:54 For the magnetic field, the magnetic field has been given as 0 .1 to 0 .i cap minus 0 .220.

03:05 J cap minus 0 .090kkk tesla this is the magnet field given so when we cross multiply 8 so i cross i will be 0 so the first term we will be getting as 0 then minus 6 .80 multiplied by 0 .2 to 0 i cap cross j cap and then in the third term this is minus 6 .80 i cap cross 0 .090 k cap so it will become first of all when we multiply this 6 .80 with the 0 .220 we will get 1 .5 and for i cross j we will get k cap and minus and here this will become plus because i cross k will come out to be minus j hence this negative with negative will become positive and the value of this will be 0 .612 for i cross k this will be j minus j and this is the v cross b now we have to find the emf induced so this will be v cross b.

04:30 So this will be v cross b.

04:33 So for v cross b we are having 0 .612 j cap minus 1 .5 k cap dot 0 .2 i cap plus 0 .15 j cap.

04:53 Here the term remaining only will be this j .j...

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