Target Exam

CUET

Subject

Physics

Chapter

Moving Charges and Magnetism

Question:

Two identical short magnetic dipoles of magnetic moment $2\, Am^2$ are placed with their axis mutually perpendicular and centers 2 m apart. The magnitude of the resultant magnetic field at the midpoint (O) between the dipoles is: ---------------

Options:

$2\sqrt{5} × 10^{-7}T$

$4\sqrt{5} × 10^{-7}T$

$2\sqrt{5} × 10^{-6}T$

$4\sqrt{5} × 10^{-6}T$

Correct Answer:

$2\sqrt{5} × 10^{-7}T$

Explanation:

The correct answer is Option (1) → $2\sqrt{5} × 10^{-7}T$

Magnetic moment of each dipole: $M = 2 \, Am^2$

Distance of midpoint from each dipole: $r = 1 \, m$

Magnetic field on axial line of dipole:

$B_{axial} = \frac{\mu_0}{4 \pi} \cdot \frac{2M}{r^3}$

$B_{axial} = \frac{10^{-7} \cdot 2 \cdot 2}{1^3} = 4 \times 10^{-7} \, T$

Magnetic field on equatorial line of dipole:

$B_{eq} = \frac{\mu_0}{4 \pi} \cdot \frac{M}{r^3}$

$B_{eq} = \frac{10^{-7} \cdot 2}{1^3} = 2 \times 10^{-7} \, T$

Since the two dipoles are perpendicular, the fields are perpendicular at midpoint.

Resultant field:

$B = \sqrt{B_{axial}^2 + B_{eq}^2}$

$B = \sqrt{(4 \times 10^{-7})^2 + (2 \times 10^{-7})^2}$

$B = \sqrt{16 + 4} \times 10^{-14}$

$B = \sqrt{20} \times 10^{-7}$

$B = 2\sqrt{5} \times 10^{-7} \, T$

Correct Answer: $2\sqrt{5} \times 10^{-7} \, T$