In the experiment to verify inverse square law with deflection Magnetometer, the value of $\frac{tan θ_{A}}{tan θ_{B}}$ will come out as :

0.25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $2$
$t a n θ_{A} = \frac{M M o 2 d}{H 4 π {(d^{2} - l^{2})}^{2}}$
$t a n θ_{B} = \frac{M M o}{H 4 π {(d^{2} + l^{2})}^{3 / 2}}$
$\frac{t a n θ_{A}}{t a n θ_{B}} = \frac{2 d {(d^{2} + l^{2})}^{3 / 2}}{{(d^{2} - l^{2})}^{2}}$

Suggest Corrections

Similar questions

In an experiment with vibration magneto-meter the value of $4 π^{2} \frac{I}{T^{2}}$ for a short bar magnet is observed as $36 \times 10^{- 4}$ . In the experiment with deflection magnetometer with the same magnet the value of $\frac{4 π d^{3}}{2 μ_{0}}$ is observed as $\frac{10^{8}}{36}$ . The magnetic moment of the magnet used is :

Q. In a magnetism experiment, the value of

\frac{m}{B}

is measured using the deflection magnetometer. What should be the value of deflection lf the percentage error is minimum?

In working with the deflection magnetometer, the proportional error will be minimum when the deflection is :

Q. The error due to imperfect centering of the pivot of the magnetic needle in the compass box in deflection magnetometer experiment is eliminated by taking the readings

In a deflection magnetometer experiment, in tan A position, a short bar Magnet placed at $18 cm$ from the centre of compass needle produces a deflection of $30^{o}$ . If another magnet of same length but 16 times pole strength as that of first magnet is placed in tan B position at $36 cm$ , the deflection will be

Join BYJU'S Learning Program

In the experiment to verify inverse square law with deflection Magnetometer, the value of tanθAtanθB will come out as :

The correct option is D 2tanθA=MMo2dH4π(d2−l2)2tanθB=MMoH4π(d2+l2)3/2tanθAtanθB=2d(d2+l2)3/2(d2−l2)2

In the experiment to verify inverse square law with deflection Magnetometer, the value of $\frac{tan θ_{A}}{tan θ_{B}}$ will come out as :

The correct option is D $2$
$t a n θ_{A} = \frac{M M o 2 d}{H 4 π {(d^{2} - l^{2})}^{2}}$
$t a n θ_{B} = \frac{M M o}{H 4 π {(d^{2} + l^{2})}^{3 / 2}}$
$\frac{t a n θ_{A}}{t a n θ_{B}} = \frac{2 d {(d^{2} + l^{2})}^{3 / 2}}{{(d^{2} - l^{2})}^{2}}$