Question

The magnetic field at the centre of a circular current carrying coil of radius R is B. The magnetic field at a distance R from the centre on the axis of the coil is

• A. $B/2$
• B. $B/\sqrt{2}$
• C. $B/2\sqrt{2}$
• D. $B/4$
• E. $B/8$

Hint:

The magnetic field is strongest at the center of the coil and drops off as you move along the axis. At a distance equal to the radius, the field strength has already dropped to about 35% of its center value.

Answer 1

The Correct Option is C

Concept:
Physics - Magnetic Field of a Circular Loop.
Step 1: State the general formula.
The magnetic field at a distance $x$ on the axis of a circular coil of radius $R$ is: $$ B_{axis} = \frac{\mu_{0} i R^{2}}{2(R^{2} + x^{2})^{3/2}} $$
Step 2: Identify the field at the centre.
At the centre, $x = 0$: $$ B = \frac{\mu_{0} i R^{2}}{2(R^{2})^{3/2}} = \frac{\mu_{0} i}{2R} $$
Step 3: Calculate the field at distance $x = R$.
Substitute $x = R$ into the general formula: $$ B' = \frac{\mu_{0} i R^{2}}{2(R^{2} + R^{2})^{3/2}} = \frac{\mu_{0} i R^{2}}{2(2R^{2})^{3/2}} $$ $$ B' = \frac{\mu_{0} i R^{2}}{2(2^{3/2} R^{3})} = \frac{\mu_{0} i}{2R \cdot 2\sqrt{2}} $$
Step 4: Relate B' to B.
Substitute the value of $B$ from Step 2: $$ B' = \frac{B}{2\sqrt{2}} $$