Question

At a point P on the axis of a current carrying circular coil of radius R, the magnetic field is B. If the distance of point P from the centre of the coil is $R\sqrt{15}$, then the magnetic field at a point Q on the axis of the coil which is at a distance of $R\sqrt{3}$ from the centre of the coil is}

• A. $B\sqrt{5}$
• B. $5B$
• C. $B\sqrt{8}$
• D. $8B$

Hint:

For axial magnetic field problems, first write \[ B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}} \] and then use ratios. Most calculations become very simple.

Answer 1

The Correct Option is B

Concept: The magnetic field on the axis of a circular current carrying coil is \[ B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}} \] where \[ R=\text{radius of coil} \] \[ x=\text{distance of the point from the centre} \] The magnetic field varies inversely as \[ (R^2+x^2)^{3/2} \] for a given current and radius.

Step 1: Write the magnetic field at point P.
Given, \[ x_P=R\sqrt{15} \] Hence, \[ B_P= \frac{\mu_0IR^2} {2(R^2+15R^2)^{3/2}} \] \[ B_P= \frac{\mu_0IR^2} {2(16R^2)^{3/2}} \] \[ B_P= \frac{\mu_0IR^2} {2(4R)^3} \] \[ B_P= \frac{\mu_0IR^2} {128R^3} \] Given that this magnetic field is equal to \(B\). \[ B_P=B \]

Step 2: Write the magnetic field at point Q.
Given, \[ x_Q=R\sqrt3 \] Therefore, \[ B_Q= \frac{\mu_0IR^2} {2(R^2+3R^2)^{3/2}} \] \[ B_Q= \frac{\mu_0IR^2} {2(4R^2)^{3/2}} \] \[ B_Q= \frac{\mu_0IR^2} {2(2R)^3} \] \[ B_Q= \frac{\mu_0IR^2} {16R^3} \]

Step 3: Find the ratio of the two magnetic fields.
\[ \frac{B_Q}{B_P} = \frac{\dfrac{\mu_0IR^2}{16R^3}} {\dfrac{\mu_0IR^2}{128R^3}} \] \[ \frac{B_Q}{B_P} = \frac{128}{16} \] \[ \frac{B_Q}{B_P}=8 \] Thus, \[ B_Q=8B \] However, simplifying directly using the standard relation: \[ B \propto \frac{1}{(R^2+x^2)^{3/2}} \] \[ \frac{B_Q}{B_P} = \left( \frac{16R^2}{4R^2} \right)^{3/2} \] \[ = 4^{3/2} \] \[ =8 \] Therefore, \[ \boxed{B_Q=8B} \]

Step 4: State the answer.
\[ \boxed{8B} \] Hence the correct option is \[ \boxed{(D)} \]