Question

Question 4.23: A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if, (a) the wire intersects the axis, (b) the wire is turned from N-S to northeast-northwest direction, (c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-171

Answer 1

Given, strength of magnetic field , B = 1.5T
radius of the cylindrical region, r = 10cm = 0.1m
current in the wire passing through the cylindrical region , I = 7A
(a) If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region.
thus, I = 2r = 0.2 m
angle between magnetic field and current ,$\theta$=90°
so, $F=Bilsin\theta$
= 1.5 × 7 × 0.2 × sin90°
= 2.1 N

(b) the wire is lowered from the axis by distance,d = 0.06m
new Length of wire , after turning it to northeast direction is given as $l_1=\frac{l}{sin\theta}$
angle between magnetic field and current,$\theta$=45°
so, force = 1.5 × 7 × 0.2/sin45° × sin45°
= 1.5 × 7 × 0.2 = 2.1 N

(c) Let I2 is the new Length of wire,
then, (l2/2)² = 4(d + r)
so, (l2)² = 4 × 4× (0.06 + 0.1) = 16 × 0.16
l2 = 1.6 m
so, force = Bil2
= 1.5 × 7 × 1.6 = 1.68N