🔢 Magnetic Effects – Numericals

Class 10 Science • Chapter 13 • Easy to Expert

📘 Numerical Practice

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Force on Conductor F = BIL sin θ
Maximum Force F = BIL
Current I = Q/t
Current from Force I = F/(BL sin θ)
🎯 Difficulty
20 TOTAL
4 EASY
8 HARD + VERY HARD
4 EXPERT
Numerical 1 – Force
EASY
Magnetic Force
A conductor of length 0.5 m carries a current of 2 A in a magnetic field of 0.4 T. The conductor is perpendicular to the field. Find the force.
आसान भाषा में: Conductor और magnetic field perpendicular हैं, इसलिए θ = 90° और sin 90° = 1 होगा।
✅ Solution
F = BIL
F = 0.4 × 2 × 0.5
F = 0.4 N
Answer: 0.4 N
Numerical 2 – Force with Current
EASY
Magnetic Force
A 0.2 m conductor carries 5 A current in a 0.3 T magnetic field at right angles. Find the force.
Current, length और magnetic field दी हुई है। 90° होने के कारण direct F = BIL use होगा।
✅ Solution
F = BIL
F = 0.3 × 5 × 0.2
Answer: 0.3 N
Numerical 3 – Current
EASY
Electric Current
A charge of 30 C passes through a conductor in 10 seconds. Calculate the current.
Charge को time से divide करके current मिलेगा।
✅ Solution
I = Q/t
I = 30/10 = 3 A
Answer: 3 A
Numerical 4 – Maximum Force
EASY
Magnetic Force
A conductor of length 0.25 m carries 4 A current in a magnetic field of 0.5 T perpendicular to the conductor. Find the maximum force.
Perpendicular होने पर force maximum होता है।
✅ Solution
F = BIL
F = 0.5 × 4 × 0.25
Answer: 0.5 N
Numerical 5 – Force at 30°
MODERATE
F = BIL sinθ
A 0.5 m conductor carries 4 A current in a 0.6 T magnetic field. The angle between them is 30°. Find the force.
इस बार conductor perpendicular नहीं है। इसलिए sin 30° = 0.5 use करेंगे।
✅ Solution
F = BIL sin θ
F = 0.6 × 4 × 0.5 × 0.5
Answer: 0.6 N
Numerical 6 – Current from Force
MODERATE
Magnetic Force
A conductor of length 0.4 m experiences a force of 0.8 N in a 0.5 T magnetic field perpendicular to it. Find current.
F = BIL में I को subject बनाइए: I = F/BL.
✅ Solution
I = F/BL
I = 0.8/(0.5 × 0.4)
I = 4 A
Answer: 4 A
Numerical 7 – Magnetic Field from Force
MODERATE
Magnetic Field
A 0.5 m conductor carrying 2 A current experiences a force of 1 N at right angles to a magnetic field. Find the magnetic field strength.
B = F/IL से magnetic field निकालेंगे।
✅ Solution
B = F/IL
B = 1/(2 × 0.5)
Answer: 1 T
Numerical 8 – Zero Force
MODERATE
Angle
A conductor carries current in a magnetic field. What is the force if the conductor is exactly parallel to the magnetic field?
Parallel होने पर θ = 0° और sin 0° = 0 होता है।
✅ Solution
F = BIL sin 0°
F = 0
Answer: 0 N
Numerical 9 – Twofold Current
HARD
Magnetic Force
A conductor experiences a force of 2 N when current is 1 A. If all other conditions remain unchanged and current becomes 3 A, what will be the new force?
F is directly proportional to I. Current 3 times होगा तो force भी 3 times होगा।
✅ Solution
F ∝ I
New F = 3 × 2
Answer: 6 N
Numerical 10 – Number of Turns Effect
HARD
Circular Loop
A circular coil produces a magnetic field of a certain strength at its centre using 10 turns. If the number of turns is doubled while current and geometry remain unchanged, how does the field strength change qualitatively?
Same current और same geometry में magnetic field number of turns के साथ increase होता है।
✅ Solution
For the same coil geometry, B ∝ N
N doubles → magnetic field doubles approximately.
Answer: Field becomes approximately twice.
Numerical 11 – Force at 60°
HARD
F = BIL sinθ
A 1 m conductor carries 2 A current in a magnetic field of 0.5 T. The angle between conductor and field is 60°. Find the force.
sin 60° = √3/2 use करेंगे।
✅ Solution
F = BIL sin 60°
F = 0.5 × 2 × 1 × √3/2
F = √3/2 N
Answer: ≈ 0.866 N
Numerical 12 – Required Current
HARD
Magnetic Force
A 0.8 m conductor is placed perpendicular to a 0.25 T magnetic field. What current is required to produce a force of 1 N?
I = F/BL.
✅ Solution
I = 1/(0.25 × 0.8)
I = 5 A
Answer: 5 A
Numerical 13 – Force Comparison
VERY HARD
Comparison
Two conductors of equal length carry currents 2 A and 5 A in the same magnetic field perpendicular to them. Find the ratio of magnetic forces on them.
Same B और L होने पर F केवल current पर depend करेगा। इसलिए force ratio = current ratio।
✅ Solution
F₁/F₂ = I₁/I₂
= 2/5
Answer: F₁ : F₂ = 2 : 5
Numerical 14 – Length Comparison
VERY HARD
Magnetic Force
Two conductors carry the same current in the same magnetic field. Their lengths are 0.2 m and 0.5 m and both are perpendicular to the field. Find the ratio of their forces.
Same B और I के लिए F directly proportional to L है।
✅ Solution
F₁/F₂ = L₁/L₂
= 0.2/0.5
Answer: F₁ : F₂ = 2 : 5
Numerical 15 – Angle Effect
VERY HARD
Angle Dependence
A conductor produces a force F when it is perpendicular to a magnetic field. What fraction of this force will act when the angle between conductor and field becomes 30°?
पहले perpendicular case में sin 90° = 1 है। अब 30° पर sin 30° = 0.5 है।
✅ Solution
F₃₀ = BIL sin 30°
F₉₀ = BIL
F₃₀/F₉₀ = 0.5
Answer: 50% of maximum force
Numerical 16 – Current Change
VERY HARD
Proportionality
A current-carrying conductor experiences a force of 4 N. If the current is increased by 50% while all other factors remain unchanged, find the new force.
50% increase का मतलब new current = 1.5 times. Force भी 1.5 times हो जाएगी।
✅ Solution
F ∝ I
F₂ = 1.5F₁
F₂ = 1.5 × 4
Answer: 6 N
Numerical 17 – Three-Factor Change
EXPERT
Magnetic Force
The magnetic field is doubled, the current is tripled and the conductor length is halved. By what factor does the force change, assuming the angle remains 90°?
F = BIL. B ×2, I ×3 और L ×1/2 है। तीनों factors multiply करें।
✅ Solution
F₂/F₁ = (2)(3)(1/2)
F₂/F₁ = 3
Answer: Force becomes 3 times.
Numerical 18 – Zero/Maximum Comparison
EXPERT
Angle
For the same conductor, current and magnetic field, compare the force when the conductor makes angles of 90°, 60° and 0° with the magnetic field.
F = BIL sinθ. सिर्फ sinθ अलग होगा: sin90 = 1, sin60 = √3/2, sin0 = 0।
✅ Solution
F₉₀ = BIL
F₆₀ = (√3/2)BIL
F₀ = 0
Answer: F₉₀ : F₆₀ : F₀ = 1 : √3/2 : 0
Numerical 19 – Required Magnetic Field
EXPERT
Multi-Step
A conductor of length 0.25 m carries 8 A current perpendicular to a magnetic field. If the force is 2 N, find the magnetic field strength.
B निकालने के लिए B = F/IL use करें। Perpendicular होने से sin90 = 1 रहेगा।
✅ Solution
B = F/IL
B = 2/(8 × 0.25)
B = 2/2
Answer: 1 T
Numerical 20 – Final Challenge
EXPERT
Mixed Magnetic Force
A 0.4 m conductor carries 5 A current in a 0.8 T field. The conductor makes an angle of 30° with the field. Calculate the force. If the angle is changed to 90°, calculate the new force and compare the two values.
पहले 30° पर sin30 = 0.5 लगाएँ। फिर 90° पर sin90 = 1 लगाकर maximum force निकालें।
✅ Step-by-Step Solution
F₃₀ = BIL sin30°
F₃₀ = 0.8 × 5 × 0.4 × 0.5
F₃₀ = 0.8 N
F₉₀ = BIL
F₉₀ = 0.8 × 5 × 0.4
F₉₀ = 1.6 N
1.6/0.8 = 2
Answer: At 30° = 0.8 N, At 90° = 1.6 N, so maximum force is twice the 30° force.
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