Question

Part B - Determine the direction of the induced current. counterclockwise It is impossible to know O clockwise Submit Request Answer Part C - Find the velocity of the bar. Templates Symbols undo redo Resetl keyboard shortcuts help V= Value Units Submit Request Answer Part D - What is the resistivity of the material the bar is made of? Templates symbols undo reqo fetet keyboard shortcuts help P= Value Units Submit Request Answer

Answer 1

We know current carrying wire experiences a magnetic force in an uniform magnetic field. The magnetic force is given by: F = I L × B ; where F, L and B are force, length vector (which is along the direction of current) and the magnetic field respectively.

Magnitude of force F = I L B Sin(Phi)

In given question, angle between L and B is 90°

So, phi = 90°

PART A : L = 60 cm = 0.6 m, I = 0.2 A, B = 2.8 T

F = I L B Sin 90° = I L B = 0.2*0.6*2.8 = 0.336 N

PART B: According to Lenz law direction of induced current is such that it reduces change in the magnetic flux

The bar is moving right. So,magnetuc field which is inward increases the magnetic flux. The induced current must be anticlockwise because this current produces its magnetic field outward the plane which reduces the increasing flux.

Direction of induced current is counterclockwise.

PART C: Induced emf = V_ind = V × B . L = V B L Sin30°

= I* Resistance = I*(R + r) = 0.2*(23+6) = 0.2*29 = 5.8

V = 5.8/(2.8*0.6*0.5) = 6.9 m/s

PART D: Given radius of the bar = diameter/2 = 0.004 m

Cross section area A = pi*(0.004)²

Resistance of the bar r = (rho)*L/A = 6

rho = 6*A/L = 6*(0.0004)²/0.6 = 1.6*10^-6 ohm.m