Share & grow the world's knowledge!
We want to connect the people who have knowledge to the people who need it, to bring together people with different perspectives so they can understand each other better, and to empower everyone to share their knowledge.
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
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
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.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
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.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
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.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
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.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/