JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A triangular shaped wire carrying \(10 A\) current is placed in a uniform magnetic field of \(0.5\,T\), as shown in figure. The magnetic force on segment \(CD\) is \(....N\) \((\) Given \(BC = CD = BD =5\,cm )\).
- A \(0.126\)
- B \(0.312\)
- C \(0.216\)
- D \(0.245\)
Answer & Solution
Correct Answer
(C) \(0.216\)
Step-by-step Solution
Detailed explanation
\(F _{ M }( CD )= BI \ell_{\text {eff }}\) \(=0.5 \times(10) \times\left(5 \sin 60 \times 10^{-2}\right)\) \(=0.216\,N\)
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