4. In Lecture 17 we saw that the net force on a current loop in a uniform magnetic field is zero. But what if B is not uniform? The diagram below shows a square loop of wire that lies in the xy-plane (Note: the +z-direction points out of the page). The loop has corners at (0,0), (0, L), (L, 0), and (L, L) and carries a current I in the clockwise direction. The loop is immersed in a magnetic field B = (Boz/L)j + (Boy/L)k, where Bo is a positive constant. y (0, L) (0, 0) (L, L) (L, 0) X What is the net magnetic force on the loop? (Hint: You will need to use the integral F = f l dl x B to find the force on each of the sides of the loop.)
4. In Lecture 17 we saw that the net force on a current loop in a uniform magnetic field is zero. But what if B is not uniform? The diagram below shows a square loop of wire that lies in the xy-plane (Note: the +z-direction points out of the page). The loop has corners at (0,0), (0, L), (L, 0), and (L, L) and carries a current I in the clockwise direction. The loop is immersed in a magnetic field B = (Boz/L)j + (Boy/L)k, where Bo is a positive constant. y (0, L) (0, 0) (L, L) (L, 0) X What is the net magnetic force on the loop? (Hint: You will need to use the integral F = f l dl x B to find the force on each of the sides of the loop.)
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![4. In Lecture 17, we saw that the net force on a current loop in a uniform magnetic field is zero. But what if \(\mathbf{B}\) is not uniform? The diagram below shows a square loop of wire that lies in the \(xy\)-plane (Note: the \(+z\)-direction points out of the page). The loop has corners at \((0,0)\), \((0,L)\), \((L,0)\), and \((L,L)\) and carries a current \(I\) in the clockwise direction. The loop is immersed in a magnetic field
\[
\mathbf{B} = (B_0 z/L) \mathbf{j} + (B_0 y/L) \mathbf{k},
\]
where \(B_0\) is a positive constant.
**Diagram Explanation:**
- The diagram shows a square loop in the \(xy\)-plane with vertices labeled: \((0,0)\), \((L,0)\), \((L,L)\), and \((0,L)\).
- The current \(I\) flows clockwise along the edges of the square loop.
**Question:**
What is the net magnetic force on the loop?
(Hint: You will need to use the integral \(\mathbf{F} = I \int d\mathbf{l} \times \mathbf{B}\) to find the force on each of the sides of the loop.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F216015d9-daff-4e98-b3d1-7f3f5202e820%2F4d965975-90a2-401e-8bc7-0f435c3d09b5%2Fnugp08_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. In Lecture 17, we saw that the net force on a current loop in a uniform magnetic field is zero. But what if \(\mathbf{B}\) is not uniform? The diagram below shows a square loop of wire that lies in the \(xy\)-plane (Note: the \(+z\)-direction points out of the page). The loop has corners at \((0,0)\), \((0,L)\), \((L,0)\), and \((L,L)\) and carries a current \(I\) in the clockwise direction. The loop is immersed in a magnetic field
\[
\mathbf{B} = (B_0 z/L) \mathbf{j} + (B_0 y/L) \mathbf{k},
\]
where \(B_0\) is a positive constant.
**Diagram Explanation:**
- The diagram shows a square loop in the \(xy\)-plane with vertices labeled: \((0,0)\), \((L,0)\), \((L,L)\), and \((0,L)\).
- The current \(I\) flows clockwise along the edges of the square loop.
**Question:**
What is the net magnetic force on the loop?
(Hint: You will need to use the integral \(\mathbf{F} = I \int d\mathbf{l} \times \mathbf{B}\) to find the force on each of the sides of the loop.)
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