A tangent galvanometer has a coil with 50 tur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a tangent galvanometer,the current $I$ is given by the formula: $I = \frac{2rB_H}{\mu_0 n} 	an 	heta$,where $r$ is the radius,$B_H$ is the h

Vedclass · Accepted Answer

$48.2$

Vedclass · Answer

(B) For a tangent galvanometer,the current $I$ is given by the formula: $I = \frac{2rB_H}{\mu_0 n} 	an 	heta$,where $r$ is the radius,$B_H$ is the horizontal component of the earth's magnetic field,$n$ is the number of turns,and $	heta$ is the deflection.Rearranging for $	an 	heta$: $	an 	heta = \frac{I \mu_0 n}{2r B_H}$.Given: $I = 0.1$ $A$,$n = 50$,$r = 4$ $cm = 0.04$ $m$,$B_H = 7 	imes 10^{-5}$ $T$,and $\mu_0 = 4\pi 	imes 10^{-7}$ $T \cdot m/A$.Substituting the values: $	an 	heta = \frac{0.1 	imes 4\pi 	imes 10^{-7} 	imes 50}{2 	imes 0.04 	imes 7 	imes 10^{-5}}$.$	an 	heta = \frac{0.1 	imes 4 	imes 3.14159 	imes 10^{-7} 	imes 50}{0.08 	imes 7 	imes 10^{-5}} = \frac{6.283 	imes 10^{-6}}{0.56 	imes 10^{-5}} = \frac{0.6283}{0.56} \approx 1.122$.Therefore,$	heta = 	an^{-1}(1.122) \approx 48.2^o$.

Similar Questions

$A$ magnet freely suspended in a vibration magnetometer makes $10$ oscillations per minute at a place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$,then its value at $B$ is

$A$ bar magnet is placed in the north-south direction with its north pole pointing towards the geographic north. In which direction from the center of the magnet will the points of zero magnetic field (neutral points) be located?

$A$ magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled,the number of oscillations made per minute would be

Vedclass Test Series

Exam Paper Generator

Online Exam Module