TS EAMCET · Physics · Magnetic Effects of Current
A square loop of length \(L\) is placed with its edges parallel to the \(X Y\)-axes. The loop is carrying the current \(I\). If the magnetic field in the region varies as \(\mathbf{B}=B_0\left(1+\frac{X y}{L^2}\right) \hat{\mathbf{k}}\), then the magnitude of the force on the loop will be
- A \(1 5IB0l L\)
- B \(\frac{f B_0 L}{2}\)
- C \(\frac{B_0 L}{\sqrt{2}}\)
- D \(\sqrt{2}: B_0 L\)
Answer & Solution
Correct Answer
(A) \(1 5IB0l L\)
Step-by-step Solution
Detailed explanation
We take one vertex of square loop at origin, for wire \(A B\), magnetic field \(A B=B_0(1+0)=B_0\). \( \therefore \quad n=0 \) For wire \(C D\), magnetic field \(=B_0\left(1+\frac{y}{L}\right)\), \( \therefore \) \( x=L \) For wire \(B C\), magnetic field…
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