Question

A magnetised needle of magnetic moment 3.6×<10-²T-¹ is placed at 30° with the direction of uniform magnetic field of magnitude 2× 10-²T calculate the torque acting on the needle​

Answer 1

Answer:

mci-xhvh-hca

Explanation:

see my smelly one fzufsifsufzigxutdigsigdufs

Answer 2

Answer:

Correct option is

B

7.2×10

−4

Nm

Given, Magnetic momentum, m=4.8×10

−2

,θ=30

0

,B=3×10

−2

T

Therefore, T=mBsinθ

=4.8×10

−2

×3×10

−2

×sin30

0

=4.8×10

−2

×3×10

−2

×

2

1

=7.2×10

−4

J

Where sin30

0

=

2

1