Question

in earth's magnetic field Bh if the frequency of oscillation of a magnetic needle is n then

Answer 1

answer : option (B) $n^2\propto B_h$

explanation : here needle is oscillate in the earth magnetic field.

so, time period of oscillation of needle in uniform earth's magnetic field is given by, $T=2\pi\sqrt{\frac{I}{MB_h}}$ ......(1)

where I is moment of inertia of needle about its axis , M is the magnetic moment of needle. and $B_h$ is earth's magnetic field.

now, relation between time period and frequency is given by,

$n=\frac{1}{T}$ .....(2)

so, frequency of needle in uniform earth's magnetic field is given by,

$n=\frac{1}{2\pi}\sqrt{\frac{MB}{I}}$

[ from equations (1) and (2), ]

here it is clear that, $n\propto\sqrt{B_h}$

or,$n^2\propto B_h$

hence, option (B) is correct

Answer 2

In earth's magnetic field Bh if the frequency of oscillation of a magnetic needle is n then n² ∝ Bh.

Explanation:

The frequency is given by the formula:

n = 1/T → (Equation 1)

Where,

T = Time period

The time period of oscillations is given as:

T = 2π √(I/(MBh)) → (Equation 2)

On substituting equation (2) in equation (1), we get,

n = 1/(2π √(I/(MBh)))

n = 1/2π × √((MBh)/I)

⇒ n ∝ √Bh

∴ n² ∝ Bh