Magnetism Problem (1 Viewer)

[MS8]

The force on the rectangular loop of wire in the magnetic field in Figure 8 can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field.

(a) What is the direction of the magnetic force on the loop? Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop.

(b) If a current of 5.00 A is used, what is the force per tesla on the 20.0-cm-wide loop?

 

[jimmysmith560]

The approach I will include requires the use of Fleming's Left-Hand rule. Since I did not take Physics, I am unsure as to whether this is within the scope of the syllabus. Here is a video to familiarise you with Fleming's Left-Hand rule (if needed):

Part a):

To determine:

The direction of magnetic force on the loop if the direction of magnetic field is out of the plane of paper and current is parallel to the ground and perpendicular to the field and justify that the forces on the sides of the loop are independent of how much the loop is in the field.

The direction of force is towards us using Fleming's Left-Hand Thumb Rule.

Explanation of solution:

Given:

The direction of field is out of the plane of paper.

The direction of current is perpendicular to field and parallel to the ground.

Formula Used:

From Fleming's Left-Hand Thumb rule: Whenever a current carrying conductor is placed inside magnetic field, a force on conductor acts on the conductor in the direction perpendicular to both conductor and magnetic field.

Calculation:

The direction of current is parallel to ground and rightwards (middle finger) and field is pointing upwards (index finger) perpendicular to the ground. Then, the thumb would point in towards us perpendicular to current and field. This is the direction of force.

The force on left side acts leftwards perpendicular to current which is parallel to the ground acting towards us (middle finger) and magnetic field which is pointing upwards outside the paper. The force on right side acts towards right perpendicular to current which is parallel to the ground acting towards away from us (middle finger) and magnetic field which is pointing upwards outside the paper. When the loop is kept in the field, equal length of sides experiences equal amount of force but in opposite direction which cancels and hence, the amount of force on the sides is independent of how much the loop is in the field and does not affect the net force.

Conclusion:

The direction of force is towards us using Fleming's Left-Hand Thumb Rule. When the loop is kept in the field, equal length of sides experiences equal amount of force but in opposite direction which cancels and hence, the amount of force on the sides is independent of how much the loop is in the field and does not affect the net force.

Part b):

To determine:

Force per tesla on 20.0 cm wide loop when a current of amount 5.00 A flows through it.

Force per tesla acting on the 20.0 cm wide loop is 1 N/T.

Explanation of solution:

Given:

Current in the loop, I = 5.00 A

Length of wire experiencing force, L = 20 cm = 0.20 m

Formula Used:

F = LIBsin𝜃

Where F is force, L is the length, I is the current and B is the magnetic field and θ is the angle between current and magnetic field.

Calculation:

F = LIBsin𝜃 ⇒ FB = 0.20 m × 5A × sin 90° = 1.0 N/T

Conclusion:

Thus, the force per tesla acting on the loop is 1.0 N/T.

I hope this helps!

 

[medaspirant]

The force on the rectangular loop of wire in the magnetic field in Figure 8 can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field.

(a) What is the direction of the magnetic force on the loop? Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop.

(b) If a current of 5.00 A is used, what is the force per tesla on the 20.0-cm-wide loop?

can u show / send pic of figure 8 cuz i think we need it to answer like what direction the field is in cuz it depends on where the north and south poles are

 

[jimmysmith560]

can u show / send pic of figure 8 cuz i think we need it to answer like what direction the field is in cuz it depends on where the north and south poles are

I believe this is Figure 8:

 

[ExtremelyBoredUser]

The approach I will include requires the use of Fleming's Left-Hand rule. Since I did not take Physics, I am unsure as to whether this is within the scope of the syllabus. Here is a video to familiarise you with Fleming's Left-Hand rule (if needed):

Part a):

To determine:

The direction of magnetic force on the loop if the direction of magnetic field is out of the plane of paper and current is parallel to the ground and perpendicular to the field and justify that the forces on the sides of the loop are independent of how much the loop is in the field.

The direction of force is towards us using Fleming's Left-Hand Thumb Rule.

Explanation of solution:

Given:

The direction of field is out of the plane of paper.

The direction of current is perpendicular to field and parallel to the ground.

Formula Used:

From Fleming's Left-Hand Thumb rule: Whenever a current carrying conductor is placed inside magnetic field, a force on conductor acts on the conductor in the direction perpendicular to both conductor and magnetic field.

Calculation:

The direction of current is parallel to ground and rightwards (middle finger) and field is pointing upwards (index finger) perpendicular to the ground. Then, the thumb would point in towards us perpendicular to current and field. This is the direction of force.

The force on left side acts leftwards perpendicular to current which is parallel to the ground acting towards us (middle finger) and magnetic field which is pointing upwards outside the paper. The force on right side acts towards right perpendicular to current which is parallel to the ground acting towards away from us (middle finger) and magnetic field which is pointing upwards outside the paper. When the loop is kept in the field, equal length of sides experiences equal amount of force but in opposite direction which cancels and hence, the amount of force on the sides is independent of how much the loop is in the field and does not affect the net force.

Conclusion:

The direction of force is towards us using Fleming's Left-Hand Thumb Rule. When the loop is kept in the field, equal length of sides experiences equal amount of force but in opposite direction which cancels and hence, the amount of force on the sides is independent of how much the loop is in the field and does not affect the net force.

Part b):

To determine:

Force per tesla on 20.0 cm wide loop when a current of amount 5.00 A flows through it.

Force per tesla acting on the 20.0 cm wide loop is 1 N/T.

Explanation of solution:

Given:

Current in the loop, I = 5.00 A

Length of wire experiencing force, L = 20 cm = 0.20 m

Formula Used:

F = LIBsin𝜃

Where F is force, L is the length, I is the current and B is the magnetic field and θ is the angle between current and magnetic field.

Calculation:

F = LIBsin𝜃 ⇒ FB = 0.20 m × 5A × sin 90° = 1.0 N/T

Conclusion:

Thus, the force per tesla acting on the loop is 1.0 N/T.

I hope this helps!

jimmy out here mastering all the hsc subjects. possible jimmy tutoring in the future??