JEE Main Important Numerical for Magnetization- Magnetism And Matter

1. 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:

$$4\times {10}^{-4}T=\frac{4\pi \times {10}^{-7}T-m/A\times 2\times 1.6A{m}^2}{(0.2m{)}^3}$$

Solving for d, we get:

$$d=0.4m$$

Therefore, the length of the magnet is 0.4 m.

2. 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:

$$B={\mu }_{0}nI$$ 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):

$$n=\frac{500turns}{0.20m}$$

$$n=2500turns/m$$

Now, we can calculate the magnetic field:

$$B=4\pi \times {10}^{-7}T-m/A\times 2500turns/m\times I$$

Assuming a current I of 1 A, we get:

$$B=4\pi \times {10}^{-7}T-m/A\times 2500turns/m\times 1A$$

$$B=3.14\times {10}^{-4}T$$

Therefore, the magnetic field inside the solenoid is 3.14 x 10^-4 T.

3. 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.

4. 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:

$$r=\frac{(1.67\times {10}^{-27}kg)({10}^{7}m/s)}{(1.6\times {10}^{-19}C)(0.1T)}$$

$$r=1.04\times {10}^{-2}m$$

Therefore, the radius of the circular path followed by the proton is 1.04 x 10^-2 m.

5. 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).

$$r=\frac{(9.11\times {10}^{-31}kg)(2\times {10}^{7}m/s)}{(-1.6\times {10}^{-19}C)(0.5T)}$$

$$r=2.82\times {10}^{-3}m$$

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

1. 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:

$$B=\frac{{\mu }_0}{4\pi }\frac{2m}{d^3}$$

$$2\times {10}^{-4}T=\frac{4\pi \times {10}^{-7}T-m/A\times 2\times 1A{m}^2}{(0.1m{)}^3}$$

$$d=0.2m$$

Therefore, the length of the magnet is 0.2 m.

2. 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:

$$B={\mu }_{0}nI$$

First, we calculate the number of turns per unit length:

$$n=\frac{100turns}{0.10m}$$

$$n=1000turns/m$$

Now, we can calculate the magnetic field:

$$B=4\pi \times {10}^{-7}T-m/A\times 1000turns/m\times I$$

Assuming a current I of 1 A:

$$B=4\pi \times {10}^{-7}T-m/A\times 1000turns/m\times 1A$$

$$B=1.26\times {10}^{-4}T$$

Therefore, the magnetic field inside the solenoid is 1.26 x 10^-4 T.

3. 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:

$$R=5cm=0.05m$$

$$B=\frac{4\pi \times {10}^{-7}T-m/A}{4\pi }\frac{2\pi \times 2.5A}{0.05m}$$

$$B=7.85\times {10}^{-4}T$$

Therefore, the magnetic field at the centre of the loop is 7.85 x 10^-4 T.

4. 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{(