Problem 27 - University Physics with Modern Physics

Question

A \(1.41-\mathrm{m}\) bar moves through a uniform, 1.20 . \(T\) magnetic field with a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) (Fig, 29.40\()\) . In cach case, find the emf induced between the ends of this bar and identify which, if any, end \((a \text { or } b)\) is at the higher potential. The bar moves in the direction of (a) the \(+x\) -axis; (b) the \(-y\) -axis; (c) the \(+z\) -axis. (d) How should this bar move so that the emf across its ends has the greatest possible value with \(b\) at a higher potential than \(a\) , and what is this maximum emf?

Step-by-Step Solution

Answer

(a) 0 V; (b) 4.23 V, b higher; (c) 4.23 V, b higher; (d) 4.23 V, direction perpendicular to the magnetic field.

Step 1: Understand the Problem and Variables

We are asked to find the induced electromotive force (emf) between the ends of a moving bar and determine which end is at a higher potential. The bar has a length of 1.41 m, moves in a magnetic field of 1.20 T, and at a speed of 2.50 m/s. We need to analyze three different movement directions: the +x-axis, -y-axis, and +z-axis.

Step 2: Use the Formula for Induced EMF

The formula for the emf induced in a bar moving in a magnetic field is \( \text{emf} = B \cdot L \cdot v \cdot \sin(\theta) \), where \( B \) is the magnetic field strength, \( L \) is the length of the bar, \( v \) is the velocity, and \( \theta \) is the angle between the direction of the velocity and the magnetic field. Since the magnetic field, assumed default by convention, is along one of the coordinate axes, judging by given cases (a, b, c), we will assume it's aligned with the +x-axis.

Step 3: Solve for Direction (a) +x-axis

For the bar moving along the +x-axis, the velocity and magnetic field are parallel. Here \( \theta = 0 \), which makes \( \sin(\theta) = 0 \). Therefore, the induced emf is 0 V, as the velocity is parallel to the magnetic field.

Step 4: Solve for Direction (b) -y-axis

For movement along the -y-axis, the velocity is perpendicular to the magnetic field (\( \theta = 90^\circ \)), thus \( \sin(\theta) = 1 \). The induced emf is calculated as: \[ \text{emf} = 1.20 \times 1.41 \times 2.50 = 4.23 \text{ V}. \] By the right-hand rule, if we assume the magnetic field is along the +x-axis, end b would be at a higher potential than end a.

Step 5: Solve for Direction (c) +z-axis

For movement along the +z-axis, the velocity is again perpendicular to the magnetic field (\( \theta = 90^\circ \)), so \( \sin(\theta) = 1 \). The induced emf is: \[ \text{emf} = 1.20 \times 1.41 \times 2.50 = 4.23 \text{ V}. \] Similar reasoning using the right-hand rule shows that it will be equivalent to case (b).

Step 6: Determine Maximum EMF and Condition (d)

To achieve the maximum emf with end b at a higher potential than end a, the velocity must be perpendicular to both the bar and the magnetic field. The maximum emf is the same as in cases where \( \theta = 90^\circ \), which is 4.23 V. Movement in any direction with angle 90 degrees to the magnetic field would achieve this emf, as shown in scenarios (b) and (c).

Key Concepts

Faraday's Law

Faraday's Law of electromagnetic induction is a fundamental principle that describes how a changing magnetic field can induce an electromotive force (emf) in a conductor. In simpler terms, when you move a conductor through a magnetic field, or the magnetic field around a conductor changes, an emf is induced, giving rise to an electric current if the circuit is closed.

The mathematical expression of Faraday's Law is given by the formula:

• \[\text{emf} = -\frac{d\Phi}{dt}\]

Here, \( \Phi \) represents the magnetic flux, and \( \frac{d\Phi}{dt} \) indicates the rate of change of magnetic flux.

In the context of our exercise, Faraday's Law helps explain why an emf is induced when the bar moves through the magnetic field. The motion of the bar changes the magnetic flux through the bar, thus inducing the emf.

magnetic fields

Magnetic fields are invisible fields that exert a force on moving charges and magnetic materials. The strength of a magnetic field is measured in teslas (T), and it defines how much force a magnetic field can exert on a moving charge.

In this exercise, a uniform magnetic field of 1.20 T is present. When the bar moves through this magnetic field, it cuts through the field lines, which is what induces an emf according to Faraday's Law. Understandably, the orientation between the motion of the conductor and magnetic field lines is crucial to determine if and how much emf will be induced.

The force experienced by charges in the bar is related to both the quantity of the magnetic field (B) and the velocity of the charge carriers (v), coupled with the length (L) of the bar moving through the field, calculated using the formula:

• \[\text{emf} = B \cdot L \cdot v \cdot \sin(\theta)\]

where \( \theta \) is the angle between the magnetic field and the direction of movement of the bar.

right-hand rule

The right-hand rule is a method used to determine the direction of the induced current or magnetic force in a scenario involving a conductor in a magnetic field. For the specific orientation of forces and fields, the right hand's thumb points in the direction of the current or velocity (the movement of a positively charged particle), and the fingers point in the direction of the magnetic field.

In our exercise, using this rule helps determine which end of the bar (end a or end b) is at higher potential. For instance, if the velocity is in a direction perpendicular to the magnetic field, say the -y-axis, and assuming the magnetic field is in the +x-axis, the palm facing direction tells us that end b will be at a higher potential due to the leftward direction of the induced emf.

induced current

Induced current is the result of an induced emf in a closed circuit. When a conductor like the bar in our exercise moves through a magnetic field and induces an emf, it can drive a current if the circuit is closed.

However, in the exercise discussed, we only consider the induced emf and not the induced current itself, as the question does not specify a complete circuit attached to the moving bar. If there were one, charges would move, creating a current that circulates through the circuit.

The induced emf calculated for different scenarios indicates the potential difference across the bar's ends, and this potential difference would drive the current once the circuit is closed, adhering to Ohm's Law: