A DMM set in tan A position. A small magnet is placed at a certain distance and the deflection observed in the needle is $53^{o}$ . The magnetic field at the site of the needle is $(B_{H} = 3.6 \times 10^{- 5} T)$

4.8 \times 10^{- 5} T

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

8 \times 10^{- 5}

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

4 \times 10^{- 5} T

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

3 \times 10^{- 5} T

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

Open in App

Solution

The correct option is A $4.8 \times 10^{- 5} T$
According to tangent law, when two uniform magnetic fields act at right angles to each other on a magnetic needle, it comes to rest in the direction of $B = B_{H} tan θ$
$= 3.6 \times 10^{- 5} \times tan 53^{\circ} T$
$= 4.8 \times 10^{- 5} T$

Suggest Corrections

Similar questions

Q. A vertical circular coil of one turn and radius 9.42 cm is placed with its plane in the magnetic meridian and short magnetic needle is pivoted at the centre of the coil so that it can freely rotate in the horizontal plane. If a current of 6A is passed through the coil, then the needle deflects by

(B_{H} = 4 \times 10^{- 5} T)

At a given place on the earths surface, the horizontal component of magnetic field is $3.2 \times 10^{- 5} T$ and the resultant magnetic field is $6.4 \times 10^{- 5} T$ . The angle of dip at that place is:

Q. Horizontal component of the earth's field is

3 \times 10^{- 5} T

and of dip is

53^{o}

. Magnetic field of the earth at that place is

Q. The vertical and horizontal component of earth's magnetic field at a place are

2 \times 10^{- 5} T

and

3.464 \times 10^{- 5} T

respectively. Calculate the angle of dip and resultant earth's magnetic field at that place.

Q. A DMM is arranged at the magnetic pole of earth in

tan A

position. If a bar magnet is placed at some distance from the needle, deflection is

Join BYJU'S Learning Program

A DMM set in tan A position. A small magnet is placed at a certain distance and the deflection observed in the needle is 53o . The magnetic field at the site of the needle is (BH=3.6×10−5T)

The correct option is A 4.8×10−5T According to tangent law, when two uniform magnetic fields act at right angles to each other on a magnetic needle, it comes to rest in the direction of B=BHtanθ=3.6×10−5×tan53∘T=4.8×10−5T

A DMM set in tan A position. A small magnet is placed at a certain distance and the deflection observed in the needle is $53^{o}$ . The magnetic field at the site of the needle is $(B_{H} = 3.6 \times 10^{- 5} T)$

The correct option is A $4.8 \times 10^{- 5} T$
According to tangent law, when two uniform magnetic fields act at right angles to each other on a magnetic needle, it comes to rest in the direction of $B = B_{H} tan θ$
$= 3.6 \times 10^{- 5} \times tan 53^{\circ} T$
$= 4.8 \times 10^{- 5} T$