A magnet of length $0.1 m$ and pole strength $10^{- 4} A m .$ is kept in a magnetic field of $30 W b / m^{2}$ at an angle $30^{o}$ . The couple acting on it is ……… $10^{- 4} N m$

7.5

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

3.0

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

1.5

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

6.0

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

Open in App

Solution

The correct option is B $6.0$
Given
Pole strength $m = 10^{- 4} A m$ and length = $0.1 m$
$τ = \to M \times \to B = M B s i n θ = m \times (2 l) \times B s i n θ = 10^{- 4} \times 0.1 \times 30 s i n 30^{o} = 1.5 \times 10^{- 4} N m$ .

Suggest Corrections

Similar questions

Q. The couple acting on a magnet of length 10 cm and pole strength 125 A-m, kept in a field of

B = 2 \times 10^{- 5} T

let at an angle of

30^{o}

Q. A bar Magnet of pole strength

2 A m

is kept in a magnetic field of induction

4 \times 10^{- 5} W b / m^{2}

such that the axis of magnet makes an angle

30^{o}

with the direction of the field. The couple acting on the magnet is found to be

80 \times 10^{- 7} N m

. Then the distance between the two poles of the magnet is :

Q. The couple acting on a bar magnet of pole strength

2 Am

when kept in a magnetic field of intensity

10 A/m

, such that axis of the magnet makes an angle

30^{o}

with the direction of the field is

80 \times 10^{- 7} Nm

. The distance between the poles of the magnet is :

A magnet of length $0.1 m$ has a pole strength of $4 A m$ . If it is making an angle $45^{o}$ with the direction of uniform magnetic field of $4 W b / m^{2}$ , the torque experienced by it is :

A bar magnet of length $10 c m$ and pole strength $2 A m$ makes an angle $60^{o}$ with a uniform magnetic field of induction $50 T$ . The couple acting on it :

Join BYJU'S Learning Program

A magnet of length 0.1m and pole strength 10−4Am. is kept in a magnetic field of 30Wb/m2 at an angle 30o. The couple acting on it is ……… 10−4Nm

The correct option is B 6.0Given Pole strength m=10−4Am and length = 0.1mτ=→M×→B=MBsinθ=m×(2l)×Bsinθ=10−4×0.1×30sin30o=1.5×10−4Nm.

A magnet of length $0.1 m$ and pole strength $10^{- 4} A m .$ is kept in a magnetic field of $30 W b / m^{2}$ at an angle $30^{o}$ . The couple acting on it is ……… $10^{- 4} N m$

The correct option is B $6.0$
Given
Pole strength $m = 10^{- 4} A m$ and length = $0.1 m$
$τ = \to M \times \to B = M B s i n θ = m \times (2 l) \times B s i n θ = 10^{- 4} \times 0.1 \times 30 s i n 30^{o} = 1.5 \times 10^{- 4} N m$ .