5.15  (a). A short bar magnet of magnetic movement 5.25 \times 10 ^{-2} JT^{-1}is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on  its normal bisector 

Answers (1)

Given,

The magnetic moment of the bar magnet, m = 5.25 \times 10^{-2}\ JT^{-1}

The magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 \times 10^{-4}   T

The magnetic field at a distance R from the centre of the magnet on the normal bisector is:

B = \mu_{0}m/4\pi R^3

When the resultant field is inclined at 45° with earth’s field, B = H 

B = H = 0.42\times 10^{-4} T

R^3 = \mu_{0}m/4\pi B  =  4\pi\times10^{-7}\times5.25\times10^{-2}/4\pi\times0.42\times10^{-4}

= 12.5 \times 10^{-5 }m^3

Therefore,  R = 0.05 m = 5 cm

Most Viewed Questions

Related Chapters

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 39999/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions