A rectangular 10.0 cm by 20.0 cm circuit carrying an 8.00 A current is oriented with its plane parallel to a uniform 0.750 T magnetic field ( Figure 20.60 ). (a) Find the magnitude and direction of the magnetic force on each segment ( ab , bc , cd , and da ) of this circuit. Illustrate your answers with clear diagrams (b) Find the magnitude of the net force on the entire circuit.
A rectangular 10.0 cm by 20.0 cm circuit carrying an 8.00 A current is oriented with its plane parallel to a uniform 0.750 T magnetic field (Figure 20.60). (a) Find the magnitude and direction of the magnetic force on each segment (ab, bc, cd, and da) of this circuit. Illustrate your answers with clear diagrams (b) Find the magnitude of the net force on the entire circuit.
Figure 20.60
(a)
The direction and magnitude magnetic force on each segment.
Answer to Problem 30P
The magnetic force is F=1.2N⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯F=1.2 N_ for segment abab, and cdcd. F→F→ is directed into the page for segment abab and out of the page for segment cdcd.
Magnetic force F=0N⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯F=0 N_ for segment bcbc and dada.
Explanation of Solution
Write the expression for magnetic force
F=IlBsinϕF=IlBsinϕ (I)
Here, FF is the force, II is the current, ll is the length of the segment, BB is the magnetic field, and ϕϕ is the angle between force and magnetic field
Conclusion:
Consider the segment abab. Segment abab is perpendicular to magnetic field thus, ϕϕ will be equal to 90°90°.
Substitute 8.0A8.0 A for II, 0.200m0.200 m for ll, 0.750T0.750 T for BB, and 90°90° for ϕϕ in equation (I)
F=8.0A×0.200m×0.750T×sin90°=1.2NF=8.0 A×0.200 m×0.750 T×sin90°=1.2 N
The direction of the magnetic field is shown below as figure1
Similarly, for segment cdcd magnetic force will be F=1.2NF=1.2 N but along abab force directed into the page and along cdcd force directed out of the page. The direction of magnetic force along cdcd is given below as figure 2
Consider segment bcbc and dada
Substitute 0°0° for ϕϕ, 8.0A8.0 A for II, 0.200m0.200 m for ll, 0.750T0.750 T for BB in equation (I)
F=8.0A×0.200m×0.750T×sin0°=0NF=8.0 A×0.200 m×0.750 T×sin0°=0 N
Therefore, the magnetic force is F=1.2N⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯F=1.2 N_ for segment abab, and cdcd. F→F→ is directed into the page for segment abab and out of the page for segment cdcd.
Magnetic force F=0N⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯F=0 N_ for segment bcbc and dada.
(b)
The net force on the circuit.
Answer to Problem 30P
The magnitude of force on the opposite segments abab, and cdcd are equal and opposite in direction, and force is zero along bcbc and dada. Thus, net force in the entire circuit will be zero.
Explanation of Solution
The magnetic force is F=1.2NF=1.2 N for segment abab, and cdcd. F→F→ is directed into the page for segment abab and out of the page for segment cdcd. Magnetic force F=0NF=0 N for segment bcbc and dada. Thus, the force along abab and cdcd will cancel off.
The net force in the entire circuit will be zero.
Conclusion:
Therefore, the net force in the entire circuit will be zero.
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