Moving Charges and Magnetism — Flashcards
Madhya Pradesh Board · Class 12 · Physics
70 flashcards for Moving Charges and Magnetism (Madhya Pradesh Board Class 12 Physics) to test yourself on key terms and facts.
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State the Lorentz force on a charge moving in electric and magnetic fields.
Answer
Formula: F = q[E(r) + v × B(r)] Meaning: The total force on a charge q is the sum of electric and magnetic parts. Units: F in newton, q in coulomb, E in N/C, v in m/s, B in tesla. Quick point: The mag…
विद्युत और चुंबकीय क्षेत्रों में गति करते एक आवेश पर Lorentz बल का कथन कीजिए।
Answer
सूत्र: F = q[E(r) + v × B(r)] अर्थ: एक आवेश q पर लगने वाला कुल बल विद्युत और चुंबकीय भागों का योग है। इकाइयाँ: F newton में, q coulomb में, E N/C में, v m/s में, B tesla में। त्वरित बिंदु: चुंबकीय भाग…
एक आवेश q = 2 C 3 m/s वेग से 4 T के चुंबकीय क्षेत्र के लम्बवत् चलता है। चुंबकीय बल ज्ञात कीजिए।
Answer
दिया है: q = 2 C, v = 3 m/s, B = 4 T, कोण = 90° सूत्र: F = qvB sinθ चरण 1: sin90° = 1 चरण 2: F = 2 × 3 × 4 × 1 N चरण 3: F = 24 N उत्तर: 24 N इकाई जाँच: C × m/s × T से N मिलता है।…
A charge q = 2 C moves with velocity 3 m/s perpendicular to a magnetic field of 4 T. Find the magnetic force.
Answer
Given: q = 2 C, v = 3 m/s, B = 4 T, angle = 90° Formula: F = qvB sinθ Step 1: sin90° = 1 Step 2: F = 2 × 3 × 4 × 1 N Step 3: F = 24 N Answer: 24 N Unit check: C × m/s × T gives N.
Why does a magnetic field do no work on a moving charge?
Answer
The magnetic force is always perpendicular to velocity. Since work done is W = F s cosθ, and θ = 90° between force and motion, W = 0. So magnetic force changes direction of motion, not speed. Example:…
एक गतिमान आवेश पर चुंबकीय क्षेत्र कोई कार्य क्यों नहीं करता?
Answer
चुंबकीय बल हमेशा वेग के लम्बवत् होता है। चूँकि किया गया कार्य W = F s cosθ है, और बल तथा गति के बीच θ = 90° है, इसलिए W = 0। इसलिए चुंबकीय बल गति की दिशा बदलता है, चाल नहीं। उदाहरण: एक आवेशित कण चुंबक…
एक आवेशित कण 5 m/s चाल से चुंबकीय क्षेत्र के समकोण पर चलता है। यदि q = 1.5 C है, तो बल ज्ञात कीजिए और वेग के साथ उसकी दिशा का संबंध बताइए।
Answer
दिया है: q = 1.5 C, v = 5 m/s, B = 2 T, θ = 90° सूत्र: F = qvB sinθ चरण 1: F = 1.5 × 5 × 2 × 1 N चरण 2: F = 15 N दिशा: बल वेग और चुंबकीय क्षेत्र दोनों के लम्बवत् है। उत्तर: 15 N…
A charged particle moves with speed 5 m/s at right angles to a magnetic field of 2 T. If q = 1.5 C, find the force and state its direction relation with velocity.
Answer
Given: q = 1.5 C, v = 5 m/s, B = 2 T, θ = 90° Formula: F = qvB sinθ Step 1: F = 1.5 × 5 × 2 × 1 N Step 2: F = 15 N Direction: The force is perpendicular to both velocity and magnetic field. Answer: 15…
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