Question

The magnetic needle of a tangent galvanometer is deflected at an angle 30 degree. The horizontal component of earths magnetic field 0.34*10^-4 t is along the plane of the coil. The magnetic field of coil

Answer 1

Answer:

The magnetic field of coil is $1.78\times10^{-5}\ T$

Explanation:

Given that,

Angle = 30°

Horizontal component of magnetic field $B_{H}=0.34\times10^{-4}$

We need to calculate the magnetic field of coil

Using formula of magnetic field

$\dfrac{B_{v}}{B_{H}}=\dfrac{B\sin\theta}{B\cos\theta}$

$\dfrac{B_{v}}{B_{H}}=\tan\theta$

Put the value into the formula

$B_{v}=\tan30\times0.34\times10^{-4}$

$B_{v}=0.0000178\ T$

$B_{v}=1.78\times10^{-5}\ T$

Hence,  The magnetic field of coil is $1.78\times10^{-5}\ T$