JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A square loop of edge length \(2 \mathrm{~m}\) carrying current of \(2 \mathrm{~A}\) is placed with its edges parallel to the \(\mathrm{x}-\mathrm{y}\) axis. A magnetic field is passing through the \(x-y\) plane and expressed as \(\vec{B}=B_0(1+4 x) \hat{k}\), where \(\mathrm{B}_0=5 \mathrm{~T}\). The net magnetic force experienced by the loop is _______ \(\mathrm{N}\).
- A \(159\)
- B \(160\)
- C \(170\)
- D \(171\)
Answer & Solution
Correct Answer
(B) \(160\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}(\mathrm{x}=0)=\mathrm{B}_0, \quad \mathrm{~B}(\mathrm{x}=2)=9 \mathrm{~B}_0\) \(\text { Also, } \mathrm{F}=\mathrm{i} \ell \mathrm{B}\) \(\Rightarrow \mathrm{F}_1=\mathrm{i} \ell \mathrm{B}_0 \& \mathrm{~F}_2=9 \mathrm{i} \ell \mathrm{B}_0\)…
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