Question

A current carrying solenoid, acting as a short bar magnet, produces a magnetic field around it. The distance of a point $\text{P}$ at its perpendicular bisector is $R$ from its centre and another point $\text{A}$ at its axis is situated at $\text{4R}$ from its centre. Then find the ratio of magnetic field at $\text{A}$ to the magnetic field at $\text{P}$

• A. $1:4$
• B. $16:1$
• C. $1:32$
• D. $1:64$

Answer 1

The correct option is C $1:32$

The magnetic field due to a solenoid acting as a bar magnet at an angle $\theta$ is given by,

$B_a=\frac{{\mu }_0}{4\pi }\frac{M}{r^3}\sqrt{3{\cos }^2\theta +1}$


As per the information given,

Field at $\text{A}:$

$B_A=\frac{{\mu }_0}{4\pi }\frac{M}{(4R{)}^3}\sqrt{3{\cos }^2{0}^{\circ }+1}=\frac{{\mu }_0}{4\pi }\frac{M}{32(R{)}^3}$

Field at point $\text{P}:$

$B_P=\frac{{\mu }_0}{4\pi }\frac{M}{R^3}\sqrt{3{\cos }^2{90}^{\circ }+1}=\frac{{\mu }_0}{4\pi }\frac{M}{R^3}$

$\therefore \frac{B_A}{B_P}=\frac{{\mu }_0}{4\pi }\frac{M}{32(R{)}^3}\times \frac{4\pi R^3}{{\mu }_0\left(M\right)}$

$\Rightarrow \frac{B_A}{B_P}=\frac{1}{32}$

Hence, $\left(C\right)$ is the correct answer.