Sankalp NEET Full Test-6 Question-3 Solution

Question 3. An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10^{5} \mathrm{~m} / \mathrm{s} parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

(1) 4 \times 10^{-20} \mathrm{~N}

(2) 8 \pi \times 10^{-20} \mathrm{~N}

(3) 4 \pi \times 10^{-20} \mathrm{~N}

(4) 8 \times 10^{-20} \mathrm{~N}

Answer: Option (4)

Explanation:

An infinitely long straight current carrying conductor produces a magnetic field around it.

The magnitude of magnetic field at a distance r from a long straight conductor carrying current I is given by:

B = \frac{\mu_0 I}{2\pi r}

Here,

I = 5 \ \mathrm{A} r = 20 \ \mathrm{cm} = 0.2 \ \mathrm{m} \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T\,m/A}

Substituting the values:

B = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.2} B = 5 \times 10^{-6} \ \mathrm{T}

The magnetic force acting on a moving charge is given by:

F = q v B \sin\theta

Here, the electron is moving parallel to the conductor,

and the magnetic field is perpendicular to the plane formed by the wire and the electron,

so \theta = 90^\circ and \sin\theta = 1.

For an electron:

q = 1.6 \times 10^{-19} \ \mathrm{C} v = 10^{5} \ \mathrm{m/s}

Therefore,

F = 1.6 \times 10^{-19} \times 10^{5} \times 5 \times 10^{-6} F = 8 \times 10^{-20} \ \mathrm{N}

Hence, the magnitude of force experienced by the electron is

8 \times 10^{-20} \ \mathrm{N}.

Thus, the correct answer is Option (4).