Q6Q In Fig. 30-26, a wire loop has b... [FREE SOLUTION]

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Chapter 30: Q6Q (page 894)

In Fig. 30-26, a wire loop has been bent so that it has three segments: segment bc(a quarter-circle), ac(a square corner), and ab(straight). Here are three choices for a magnetic field through the loop:
$(1) \overset{鈫�}{B_{1}} = 3 \hat{i} + 7 \hat{j} - 5 t \hat{k}, (2) \overset{鈫�}{B_{2}} = 5 t \hat{i} - 4 \hat{j} - 15 \hat{k}, (3) \overset{鈫�}{B_{3}} = 2 \hat{i} - 5 t \hat{j} - 12 \hat{k},$
where Bis in milliteslas and tis in seconds. Without written calculation, rank the choices according to (a) the work done per unit charge in setting up the induced current and (b) that induced current, greatest first. (c) For each choice, what is the direction of the induced current in the figure?

Short Answer

Expert verified

The rank of magnetic fields according to the work done per unit charge in setting up the induced current: $\overset{鈫�}{B_{2}}, \overset{鈫�}{B_{1}}, \overset{鈫�}{B_{3}}$ .
The rank of magnetic fields according to the induced current: $\overset{鈫�}{B_{2}}, \overset{鈫�}{B_{1}}, \overset{鈫�}{B_{3}}$ .
The direction of each induced current will be: $\overset{鈫�}{B_{2}}$ : counter-clockwise, $\overset{鈫�}{B_{1}}$ : clockwise, and $\overset{鈫�}{B_{3}}$ : counterclockwise.

Step by step solution

Step 1: Given

The three segments of the loop are bc (a quarter circle), ac (a square corner), and ab (straight).
The three choices of a magnetic field are,

$\overset{鈫�}{B_{1}} = 3 \hat{i} + 7 \hat{j} - 5 t \hat{k}, \overset{鈫�}{B_{2}} = 5 t \hat{i} - 4 \hat{j} - 15 \hat{k}, \overset{鈫�}{B_{3}} = 2 \hat{i} - 5 t \hat{j} - 12 \hat{k},$

Determining the concept

The magnitude of the induced emf is the work done per unit charge in setting up the induced current. It depends upon the magnitude of the changing magnetic flux through the area of the loop. It also depends upon the angle between the magnetic field and the area vector for each part of the loop.

$蔚 = \frac{鈥� d 蠒}{d t} 蠒 = \overset{鈫�}{B} . \overset{鈫�}{A}$

Where,

$d 蠒$ = change in magnetic flux,

dt= change in time,

B = magnetic field,

A = area,

$蠒$ = magnetic flux,

(a) Determining the rank of magnetic fields according to the work done per unit charge in setting up the induced current

According to Faraday鈥檚 law, changing magnetic flux induces current and emf in the loop. The induced emf is the work done per unit charge in setting up the induced current. The magnitude of magnetic flux also depends upon the angle between the magnetic field and area vector, given as

$蔚 = \frac{鈥� d 蠒}{d t} 蠒 = \overset{鈫�}{B} . \overset{鈫�}{A}$

Comparing the areas of all sections of the loop, find the area of the loop by segment ac is the greatest, the area of the loop by segment bc is smaller than that and the area of the loop by the segment ab is the least.

By observing the choices of the magnetic fields, the changing magnetic flux will be maximum for the segment ac for choice role="math" localid="1661421114929" $\overset{鈫�}{B_{2}}$ . This is because the X component of the $\overset{鈫�}{B_{2}}$ is changing with time while other components are constant with time. Also, this time-varying component $\overset{鈫�}{B_{2}}$ is perpendicular to segment ac, which has the greatest area. Thus emf i.e. work done per unit charge in setting up the induced current will be maximum.

For choice $\overset{鈫�}{B_{1}}$ , the time-varying component is the Z component. This is perpendicular to the segment bc. Hence, the changing magnetic flux will be smaller than that for $\overset{鈫�}{B_{2}}$ .

For choice $\overset{鈫�}{B_{3}}$ , the time-varying component is the Y component. This is perpendicular to the segment ab. Hence the changing magnetic flux will be the least $\overset{鈫�}{B_{3}}$ among the three.

Hence, the rank of magnetic fields according to the work done per unit charge in setting up the induced current: $\overset{鈫�}{B_{2}}, \overset{鈫�}{B_{1}}, \overset{鈫�}{B_{3}}$ .

(b) Determining the rank of magnetic fields according to the induced current

Since the magnetic flux change is maximum for $\overset{鈫�}{B_{2}}$ , the magnitude of induced emf will also be maximum for $\overset{鈫�}{B_{2}}$ . For $\overset{鈫�}{B_{1}}$ , the magnitude of induced current will be smaller than that for $\overset{鈫�}{B_{2}}$ and the induced current will be least for $\overset{鈫�}{B_{3}}$ .

Hence, the rank of magnetic fields according to the induced current: $\overset{鈫�}{B_{2}}, \overset{鈫�}{B_{1}}, \overset{鈫�}{B_{3}}$ .

(c) Determining the direction of each induced current

For $\overset{鈫�}{B_{1}}$ , the direction of the time-varying magnetic field is along 鈥�Z-axis. Hence, the direction of the induced current should be such that it creates a field that opposes this field as per the Lenz law. Therefore, the current will be counterclockwise.

Similarly, $\overset{鈫�}{B_{2}}$ , the direction of the time-varying magnetic field is along the +X axis. Hence the direction of the induced current should be such that it creates a field that opposes this field as per the Lenz law. Hence, the current will be clockwise.

Similarly, $\overset{鈫�}{B_{3}}$ , the direction of the time-varying magnetic field is along -Y-axis. Hence, it will be counterclockwise. Therefore, the current will be counter-clockwise.

Hence, the direction of each induced current will be:

$\overset{鈫�}{B_{2}}$ : counter-clockwise, $\overset{鈫�}{B_{1}}$ : clockwise, and $\overset{鈫�}{B_{3}}$ : counterclockwise.

The work done per unit charge is the emf induced in the loop. The magnitude of induced emf is decided by the magnitude of the changing magnetic flux through the area of the loop i.e. by the magnitude of the field and the angle between the area vector and magnetic field. By comparing the magnitudes of the areas and the magnetic flux, we can rank the choices of the magnetic field.