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JEE Main Important Numerical for Magnetization- Magnetism And Matter
- A bar magnet has a magnetic moment of 1.6 Am^2. The magnetic field at a point 20 cm from the centre of the magnet on its axial line is 4 x 10^-4 T. Calculate the length of the magnet.
Solution: Using the formula for the magnetic field at a point on the axial line of a bar magnet:
$$B=\frac{\mu_0}{4\pi}\frac{2m}{d^3}$$ Where:
- B is the magnetic field (4 x 10^-4 T)
- μ0 is the vacuum permeability (4π x 10^-7 T-m/A)
- m is the magnetic moment (1.6 Am^2)
- d is the distance from the centre of the magnet to the point (20 cm or 0.2 m)
Substituting the given values, we get:
$$4\times10^{-4} T = \frac{4\pi\times 10^{-7}T-m/A\times 2\times 1.6Am^2}{(0.2m)^3}$$
Solving for d, we get:
$$d = 0.4 m$$
Therefore, the length of the magnet is 0.4 m.
- A solenoid 20 cm long and 5 cm in diameter has 500 turns. Calculate the magnetic field inside the solenoid.
Solution: Using the formula for the magnetic field inside a solenoid:
$$B=\mu_0nI$$ Where:
- B is the magnetic field
- μ0 is the vacuum permeability (4π x 10^-7 T-m/A)
- n is the number of turns per unit length
- I is the current flowing through the solenoid
First, we need to calculate the number of turns per unit length (n):
$$n = \frac{500\ turns}{0.20m}$$
$$n = 2500\ turns/m$$
Now, we can calculate the magnetic field:
$$B = 4\pi\times 10^{-7} T-m/A \times 2500\ turns/m\times I$$
Assuming a current I of 1 A, we get:
$$B = 4\pi\times 10^{-7}T-m/A \times 2500\ turns/m\times 1A$$
$$B = 3.14\times 10^{-4} T$$
Therefore, the magnetic field inside the solenoid is 3.14 x 10^-4 T.
- A current-carrying wire is bent into a circular loop of radius 10 cm. The current flowing through the wire is 5 A. Calculate the magnetic field at the centre of the loop.
Solution: Using the formula for the magnetic field at the centre of a circular loop:
$$B = \frac{\mu_0}{4\pi}\frac{2\pi I}{R}$$
Where:
- B is the magnetic field
- μ0 is the vacuum permeability (4π x 10^-7 T-m/A)
- I is the current flowing through the loop (5 A)
- R is the radius of the loop (10 cm or 0.1 m)
Substituting the given values, we get:
$$B = \frac{4\pi\times 10^{-7} T-m/A}{4\pi}\frac{2\pi \times 5A}{0.1m}$$
$$B = 2\times 10^{-4} T$$
Therefore, the magnetic field at the centre of the loop is 2 x 10^-4 T.
- A proton moving with a speed of 10^7 m/s enters a magnetic field of 0.1 T perpendicular to its velocity. Calculate the radius of the circular path followed by the proton.
Solution: Using the formula for the radius of the circular path followed by a charged particle in a magnetic field:
$$r = \frac{mv}{qB}$$
Where:
- r is the radius of the circular path
- m is the mass of the particle (proton’s mass = 1.67 x 10^-27 kg)
- v is the velocity of the particle (10^7 m/s)
- q is the charge of the particle (proton’s charge = 1.6 x 10^-19 C)
- B is the magnetic field strength (0.1 T)
Substituting the given values, we get:
$$r = \frac{(1.67\times 10^{-27}kg)(10^7 m/s)}{(1.6\times 10^{-19}C)(0.1 T)}$$
$$r = 1.04\times 10^{-2} m$$
Therefore, the radius of the circular path followed by the proton is 1.04 x 10^-2 m.
- An electron moving with a speed of 2 x 10^7 m/s enters a magnetic field of 0.5 T perpendicular to its velocity. Calculate the radius of the circular path followed by the electron.
Solution: Following the same formula as before:
$$r = \frac{mv}{qB}$$
But in this case, we’ll use the electron’s mass (9.11 x 10^-31 kg) and charge (-1.6 x 10^-19 C).
$$r = \frac{(9.11\times 10^{-31}kg)(2\times 10^7 m/s)}{(-1.6\times 10^{-19}C)(0.5 T)}$$
$$r = 2.82\times 10^{-3} m$$
Therefore, the radius of the circular path followed by the electron is 2.82 x 10^-3 m.
CBSE Board Exam Important Numerical for Magnetization- Magnetism And Matter
- A bar magnet has a magnetic moment of 1 Am^2. The magnetic field at a point 10 cm from the centre of the magnet on its axial line is 2 x 10^-4 T. Calculate the length of the magnet.
Solution: Using the same formula as before, but with the given values:
$$B=\frac{\mu_0}{4\pi}\frac{2m}{d^3}$$
$$2\times 10^{-4}T = \frac{4\pi\times 10^{-7} T-m/A\times 2\times 1Am^2}{(0.1m)^3}$$
$$d = 0.2 m$$
Therefore, the length of the magnet is 0.2 m.
- A solenoid 10 cm long and 2.5 cm in diameter has 100 turns. Calculate the magnetic field inside the solenoid.
Solution: Again, using the formula for the magnetic field inside a solenoid:
$$B=\mu_0nI$$
First, we calculate the number of turns per unit length:
$$n = \frac{100\ turns}{0.10m}$$
$$n = 1000\ turns/m$$
Now, we can calculate the magnetic field:
$$B = 4\pi\times 10^{-7} T-m/A \times 1000\ turns/m\times I$$
Assuming a current I of 1 A:
$$B = 4\pi\times 10^{-7} T-m/A \times 1000\ turns/m\times 1A$$
$$B = 1.26\times 10^{-4} T$$
Therefore, the magnetic field inside the solenoid is 1.26 x 10^-4 T.
- A current-carrying wire is bent into a circular loop of radius 5 cm. The current flowing through the wire is 2.5 A. Calculate the magnetic field at the centre of the loop.
Solution: Following the formula:
$$B = \frac{\mu_0}{4\pi}\frac{2\pi I}{R}$$
Where:
$$R= 5 cm = 0.05m$$
$$B = \frac{4\pi\times 10^{-7} T-m/A}{4\pi}\frac{2\pi \times 2.5A}{0.05m}$$
$$B = 7.85\times 10^{-4} T$$
Therefore, the magnetic field at the centre of the loop is 7.85 x 10^-4 T.
- A proton moving with a speed of 5 x 10^6 m/s enters a magnetic field of 0.05 T perpendicular to its velocity. Calculate the radius of the circular path followed by the proton.
Solution:
$$r = \frac{mv}{qB}$$
$$r = \frac{(
