Question

A simple pendulum with a bob of mass m and conducting wire of length L swings under gravity through an angle θ. The component of the earth's magnetic field in the direction perpendicular to the swing is B. Maximum emf induced across the pendulum is ______.

(g = acceleration due to gravity)

 

विकल्प

• `BL sin(theta/2)(gL)`

• `2BLsin(theta/2)(gL)^{3"/"2}`

• `2BLsin(theta/2)(gL)^2`

• `2BLsin(theta/2)(gL)^{1"/"2}`

Answer 1

A simple pendulum with a bob of mass m and conducting wire of length L swings under gravity through an angle θ. The component of the earth's magnetic field in the direction perpendicular to the swing is B. Maximum emf induced across the pendulum is `underlinebb(2BLsin(theta/2)(gL)^{1"/"2})`.

Explanation:

Height of the bob at max angle:
h = L - L cos θ = L(1 - cosθ)

Conversion to half-angle identity:
h = L(2sin²(`theta/2`))

Conservation of energy at the lowest point (max velocity):
mgh = `1/2` mv²_maxgL(2sin²(`theta/2`)) = `1/2`v²_max

Solve for maximum velocity:

v²_max = 4gLsin²(`theta/2`)vmax = `sqrt(4gLsin^2(theta/2)) = 2sin(theta/2)sqrt(gL)`

Maximum induced EMF (Motional EMF):

E_max = BLv_max

Substitute v_max into EMF equation:

E_max = BL (2sin (`theta/2`)`sqrt(gL)`)E_max = 2BLsin(`theta/2`)(gL)^1/2