Let the length of each wire be L. For a square of side a, 4a = L ⟹ a = L/4. Magnetic moment, M₁ = IA² = I L/4 ². For a circle of radius r, 2πr = L ⟹ r = L/2π. Magnetic moment, M₂ =Iπr² = Iπ L/2π ² . Ration. M₁/M₂ = 2/π. Answer: (A) 2 : π For more visit here: https://www.tiwariacademy.com/ncert-solutRead more
Let the length of each wire be L.
For a square of side a, 4a = L ⟹ a = L/4. Magnetic moment,
M₁ = IA² = I L/4 ².
For a circle of radius r, 2πr = L ⟹ r = L/2π. Magnetic moment,
M₂ =Iπr² = Iπ L/2π ² .
Ration. M₁/M₂ = 2/π.
Answer: (A) 2 : π
Current sensitivity of a galvanometer is given by S = nBA/K where n = number of turns, B = magnetic field, A = area of coil and k = torsional constant. To increase sensitivity, S should be maximized. Decreasing the torsional constant k increases S as they are inversely proportional. Hence, the correRead more
Current sensitivity of a galvanometer is given by
S = nBA/K
where n = number of turns, B = magnetic field, A = area of coil and k = torsional constant.
To increase sensitivity, S should be maximized. Decreasing the torsional constant
k increases S as they are inversely proportional.
Hence, the correct answer is (C) Torsional constant k.
The magnetic field due to a long straight current-carrying wire is given by: B = μ0I/2πr Here, B ∝ 1/r . If the distance increases from 5 cm to 20 cm (i.e., four times), the magnetic field decreases by the same factor: B′ = B/4. Therefore, the new magnetic field is B/4. Answer: (B) B/4. For more visRead more
The magnetic field due to a long straight current-carrying wire is given by:
B = μ0I/2πr
Here, B ∝ 1/r .
If the distance increases from 5 cm to 20 cm (i.e., four times), the magnetic field decreases by the same factor:
B′ = B/4.
Therefore, the new magnetic field is B/4.
Answer: (B) B/4.
Magnetic induction at the center of a single-turn circular loop carrying current I is given by: B = μ0I/2R When the wire is looped into two turns, the number of turns n = 2 and the magnetic induction becomes: B' = n² B = 2²B = 4B Thus, the new magnetic induction at the center is 4B. Answer: (C) 4BRead more
Magnetic induction at the center of a single-turn circular loop carrying current I is given by:
B = μ0I/2R
When the wire is looped into two turns, the number of turns
n = 2 and the magnetic induction becomes:
B’ = n² B = 2²B = 4B
Thus, the new magnetic induction at the center is 4B.
Answer: (C) 4B.
The magnetic force per unit length between two parallel currents is given by: F/L = μ0I₁ I₂/2πd Here, I₁ = 10 A, I₂ = 2 A, d = 0.1 m: Substituting values: F/L = (4π ×10⁻⁷) (10) (2)/2π (0.1) = 4 × 10⁻⁵ N/m For 2 m length of wire B, the total force is: F = 4 × 10⁻⁵ × 2 = 8 × 10⁻⁵ N Answer: (A) 8 × 1Read more
The magnetic force per unit length between two parallel currents is given by:
F/L = μ0I₁ I₂/2πd
Here, I₁ = 10 A, I₂ = 2 A, d = 0.1 m: Substituting values:
F/L = (4π ×10⁻⁷)
(10) (2)/2π (0.1) = 4 × 10⁻⁵ N/m
For 2 m length of wire B, the total force is:
F = 4 × 10⁻⁵ × 2 = 8 × 10⁻⁵ N
Answer: (A) 8 × 10⁻⁵ N.
Two wires of same length are shaped into a squatr and a circle if they carry same current, ratio of magnetic moment is :
Let the length of each wire be L. For a square of side a, 4a = L ⟹ a = L/4. Magnetic moment, M₁ = IA² = I L/4 ². For a circle of radius r, 2πr = L ⟹ r = L/2π. Magnetic moment, M₂ =Iπr² = Iπ L/2π ² . Ration. M₁/M₂ = 2/π. Answer: (A) 2 : π For more visit here: https://www.tiwariacademy.com/ncert-solutRead more
Let the length of each wire be L.
For a square of side a, 4a = L ⟹ a = L/4. Magnetic moment,
M₁ = IA² = I L/4 ².
For a circle of radius r, 2πr = L ⟹ r = L/2π. Magnetic moment,
M₂ =Iπr² = Iπ L/2π ² .
Ration. M₁/M₂ = 2/π.
Answer: (A) 2 : π
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Current sensitivity of a galvanometer can be increased by decreasing:
Current sensitivity of a galvanometer is given by S = nBA/K where n = number of turns, B = magnetic field, A = area of coil and k = torsional constant. To increase sensitivity, S should be maximized. Decreasing the torsional constant k increases S as they are inversely proportional. Hence, the correRead more
Current sensitivity of a galvanometer is given by
S = nBA/K
where n = number of turns, B = magnetic field, A = area of coil and k = torsional constant.
To increase sensitivity, S should be maximized. Decreasing the torsional constant
k increases S as they are inversely proportional.
Hence, the correct answer is (C) Torsional constant k.
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An electric current passes through a long straight copper wire. At a perpendicular distance 5 cm from the straight wire, the magnetic field is B. The magnetic field at perpendicular distance 20 cm from the straight wire would be
The magnetic field due to a long straight current-carrying wire is given by: B = μ0I/2πr Here, B ∝ 1/r . If the distance increases from 5 cm to 20 cm (i.e., four times), the magnetic field decreases by the same factor: B′ = B/4. Therefore, the new magnetic field is B/4. Answer: (B) B/4. For more visRead more
The magnetic field due to a long straight current-carrying wire is given by:
B = μ0I/2πr
Here, B ∝ 1/r .
If the distance increases from 5 cm to 20 cm (i.e., four times), the magnetic field decreases by the same factor:
B′ = B/4.
Therefore, the new magnetic field is B/4.
Answer: (B) B/4.
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A wire in the form of a circular loop, of one turn carrying a current, produces magnetic induction B at the centre. If the same wire is looped into a coil of two turns and carries the same current, the new value of magnetic induction at the centre is
Magnetic induction at the center of a single-turn circular loop carrying current I is given by: B = μ0I/2R When the wire is looped into two turns, the number of turns n = 2 and the magnetic induction becomes: B' = n² B = 2²B = 4B Thus, the new magnetic induction at the center is 4B. Answer: (C) 4BRead more
Magnetic induction at the center of a single-turn circular loop carrying current I is given by:
B = μ0I/2R
When the wire is looped into two turns, the number of turns
n = 2 and the magnetic induction becomes:
B’ = n² B = 2²B = 4B
Thus, the new magnetic induction at the center is 4B.
Answer: (C) 4B.
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Currents of 10 A and 2 A are flowing in opposite directions through two parallel wires A and B respectively. If the wire A is infinitely long and wire B is 2 m long, then force on wire B which is situated at 10 cm from A, is
The magnetic force per unit length between two parallel currents is given by: F/L = μ0I₁ I₂/2πd Here, I₁ = 10 A, I₂ = 2 A, d = 0.1 m: Substituting values: F/L = (4π ×10⁻⁷) (10) (2)/2π (0.1) = 4 × 10⁻⁵ N/m For 2 m length of wire B, the total force is: F = 4 × 10⁻⁵ × 2 = 8 × 10⁻⁵ N Answer: (A) 8 × 1Read more
The magnetic force per unit length between two parallel currents is given by:
F/L = μ0I₁ I₂/2πd
Here, I₁ = 10 A, I₂ = 2 A, d = 0.1 m: Substituting values:
F/L = (4π ×10⁻⁷)
(10) (2)/2π (0.1) = 4 × 10⁻⁵ N/m
For 2 m length of wire B, the total force is:
F = 4 × 10⁻⁵ × 2 = 8 × 10⁻⁵ N
Answer: (A) 8 × 10⁻⁵ N.
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